Samacheer Kalvi Class 11 Bio Botany Solutions Chapter 8 Biomolecules

Part I

 

Question 1. The most basic amino acid is
(a) Arginine
(b) Histidine
(c) Glycine
(d) Glutamine
Answer: (a) Arginine
In simple words: Arginine is an amino acid that has a very basic side chain, meaning it contains a nitrogen atom that can easily accept a proton, making it the most basic. This basic nature is important for how it interacts with other molecules in the body.

🎯 Exam Tip: Remember the basic amino acids (Arginine, Lysine, Histidine) and their structural features to easily identify them.

 

Question 2. An example of feed back inhibition is
(a) cyanide action on cytochrome
(b) Sulpha drug on folic acid
(c) Allosteric inhibition of hexokinase by glucose-6-phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer: (c) Allosteric inhibition of hexokinase by glucose-6-phosphate
In simple words: Feedback inhibition is like a control system where the final product of a pathway slows down the first step. Here, when enough glucose-6-phosphate is made, it tells hexokinase to slow down, stopping more glucose from being used up unnecessarily. This helps the cell save energy and resources by not making too much of something it already has.

🎯 Exam Tip: Understand that feedback inhibition is a crucial regulatory mechanism in metabolism, ensuring efficiency and balance in biochemical pathways.

 

Question 3. Enzymes that catalyse interconversion of optical, geometrical or positional isomers are
(a) Ligases
(b) Lyases
(c) Hydrolases
(d) Isomerases
Answer: (d) Isomerases
In simple words: Isomerases are a special type of enzyme that change one molecule into another form of the same molecule, called an isomer. They help in rearranging the atoms within a molecule to create different shapes or positions without adding or removing any parts. This is like turning a right-handed glove into a left-handed one, without changing the material.

🎯 Exam Tip: Recognize the function of each enzyme class; isomerases are unique for their role in converting between different isomeric forms of a compound.

 

Question 4. Proteins perform many physiological functions, for example, some functions as enzymes one of the following represents an additional function that some proteins discharge
(a) Pigment conferring colour to skin
(b) Pigments making colours of flowers
(c) Hormones
(d) None of the options
Answer: (c) Hormones
In simple words: Besides working as enzymes, some proteins also act as hormones. Hormones are chemical messengers that help control many body functions, like growth and metabolism. Insulin, for example, is a protein hormone that helps regulate blood sugar levels.

🎯 Exam Tip: Remember that proteins are highly versatile molecules, serving diverse roles beyond structural support and enzymatic activity, including signaling as hormones.

 

Question 5. Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown & one Blank component 'X' in it.

Answer: (IV) Nucleoside
In simple words: The diagram shows a sugar molecule linked to a nitrogenous base, which together form a nucleoside. The blank component 'X' would represent the nitrogenous base, such as Uracil, which is a key part of genetic material like RNA.

🎯 Exam Tip: Be able to recognize the basic structures of biomolecules like nucleosides, nucleotides, amino acids, and their components for identification questions.

 

Question 6. Distinguish between Nitrogen base and a base found in inorganic chemistry
Answer:

In simple words: Nitrogenous bases are organic molecules, essential for DNA and RNA, containing nitrogen atoms that can accept protons. Inorganic bases are simple compounds, usually metal oxides or hydroxides, that are bitter, slippery, and turn red litmus blue. The main difference is their structure and role in living things versus non-living chemistry.

🎯 Exam Tip: When distinguishing between two concepts, always define each clearly and then highlight key differences in structure, function, and examples.

 

Question 7. What are the factors affecting the rate of enzyme reactions?
Answer: Enzymes are biological molecules that are sensitive to their surroundings. Several factors can influence how fast they work.
(i) Temperature

• Heating increases the movement of molecules, which quickens the enzyme reaction up to a certain point.
• There is an ideal temperature, called the optimum temperature, where the enzyme works best and promotes maximum activity.

(ii) pH

• Changes in pH can alter the shape of an enzyme's active site, which is the part where the reaction happens.
• Extreme pH levels can permanently damage enzymes, a process called denaturation.
• The optimum pH is the specific pH level at which the enzyme's reaction rate is highest.

(iii) Substrate concentration

• For a given amount of enzyme, the reaction rate increases as more substrate is added. This is because there are more molecules for the enzyme to work on.

(iv) Enzyme concentration

• The rate of a reaction is directly related to how much enzyme is present. More enzymes mean more active sites are available, leading to a faster reaction.

In simple words: Enzyme reactions are affected by temperature, pH, and the amounts of both the enzyme and the substance it acts on. Each enzyme has an ideal temperature and pH where it works fastest, and having more enzyme or more substrate can speed up the reaction.

🎯 Exam Tip: Remember that enzymes are proteins, and their activity is highly dependent on their three-dimensional structure, which is sensitive to environmental changes like pH and temperature.

 

Question 8. Briefly outline the classification of enzymes?
Answer: Enzymes are classified into six main types based on the kind of reactions they catalyze. Each class performs a specific type of chemical transformation in the body.

In simple words: Enzymes are categorized into six main groups based on the chemical changes they cause. These groups include oxidoreductases (for redox reactions), transferases (for moving groups), hydrolases (for breaking bonds with water), isomerases (for rearranging atoms), lyases (for breaking bonds without water), and ligases (for forming new bonds using energy).

🎯 Exam Tip: Memorize the six main classes of enzymes and the general type of reaction each class catalyzes, along with a common example for each.

 

Question 9. Write down the characteristic features of DNA?
Answer: Deoxyribonucleic acid (DNA) has several unique features that make it the carrier of genetic information. Its double helix structure is key to its function.

1. If one strand runs in the 5′ – 3' direction, the other runs in the 3′ – 5' direction. This means they are antiparallel, running in opposite directions. The 5' end has a phosphate group, and the 3' end has an OH (hydroxyl) group.
2. The angle at which the two sugars stick out from the base pairs is about 120 degrees for the narrow angle and 240 degrees for the wide angle. This creates a minor groove and a large angle (major groove) between the sugars on the other edge of the helix.
3. Each base is 0.34 nm apart. A complete turn of the helix in the most common B form of DNA contains 3.4 nm and 10 base pairs per turn.
4. The DNA helical structure has a diameter of 20 Å (angstroms) and a pitch (the length of one complete turn) of about 3 Å. X-ray studies show that it takes about 10 base pairs to complete one full turn of the helix (360 degrees).
5. The stability of the DNA helix and the specific pairing of bases come from:
   • The hydrogen bonds between complementary bases (A with T, G with C) in the double helix.
   • The stacking interaction where bases sit on top of each other, roughly perpendicular to the helical axis, providing additional stability.
6. Electron cloud interactions (pi-pi stacking interactions) between the bases within the helical stacks also contribute to the stability of the double helix.
7. The phosphodiester linkages, which connect the sugar and phosphate groups, give the DNA helix a built-in direction (polarity). These linkages form strong covalent bonds, giving strength and stability to the polynucleotide chain.
8. Plectonemic coiling means the two DNA strands are twisted around each other in a helix, making it impossible to separate them without breaking the entire structure. In contrast, paranemic coiling, where strands lie side-by-side, makes them easier to separate.
9. Based on its helical structure and the distance between turns, DNA exists in three main forms: A-DNA, B-DNA, and Z-DNA.

In simple words: DNA has two strands that run in opposite directions and twist into a double helix shape. This shape is very stable due to hydrogen bonds between base pairs and stacking interactions. Each turn of the helix has 10 base pairs, and the whole molecule has a specific direction because of its sugar-phosphate backbone.

🎯 Exam Tip: Focus on understanding the double helix structure, antiparallel strands, base pairing rules (Chargaff's rules), and the role of hydrogen bonds in maintaining stability.

 

Question 10. Explain the structure and function of different types of RNA?
Answer: Ribonucleic acid (RNA) comes in several types, each playing a critical role in gene expression and protein synthesis. Unlike DNA, RNA is typically single-stranded and can fold into complex 3D shapes.

1. **mRNA (messenger RNA)**
   • Is single-stranded.
   • Carries a copy of genetic instructions from DNA to guide amino acid assembly for protein synthesis.
   • Is generally unstable, meaning it has a short lifespan.
   • Makes up about 5% of the total RNA in a cell.
   • In prokaryotes, it is polycistronic, meaning it carries coding sequences for many proteins.
   • In eukaryotes, it is monocistronic, carrying information for only one protein.
2. **tRNA (transfer RNA)**
   • Is a single-stranded molecule that folds into a clover-shaped structure with four highly folded arms.
   • Its job is to translate the genetic code from mRNA and transfer specific amino acids to the ribosomes for protein building.
   • Is unstable, also known as soluble RNA.
   • Makes up about 15% of the total RNA.
3. **rRNA (ribosomal RNA)**
   • Is single-stranded.
   • Forms the two subunits of ribosomes, which are the cell's protein-making factories.
   • Is metabolically stable, meaning it lasts longer in the cell.
   • Makes up about 80% of the total RNA.
   • Is a polymer with varying lengths, from 120 to 3000 nucleotides, and gives ribosomes their specific shape.
   • The genes for rRNA are often used in phylogenetic studies to understand evolutionary relationships.

In simple words: There are three main types of RNA: mRNA, tRNA, and rRNA. mRNA carries instructions from DNA to make proteins. tRNA helps bring the right building blocks (amino acids) to the protein-making machine. rRNA is a main part of the protein-making machine itself, called the ribosome.

🎯 Exam Tip: Clearly distinguish between the structure, percentage in the cell, and primary function of mRNA, tRNA, and rRNA. Focus on their roles in the central dogma of molecular biology.

I Choose The Right Answer.

 

Question 1. Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer: (c) Knoll and Ruska
In simple words: The electron microscope, which allows us to see very tiny details, was first invented by scientists named Knoll and Ruska. This invention greatly helped in studying very small things like cells and viruses.

🎯 Exam Tip: For scientific discoveries, it's important to remember the key scientists associated with major inventions or theories.

 

Question 2. Polysaccharides are also called
(a) Polymers
(b) Glycans
(c) Glycosidic compounds
(d) Glycones
Answer: (b) Glycans
In simple words: Polysaccharides are large sugar molecules made of many smaller sugar units linked together. They are also known as glycans, a term that specifically refers to these carbohydrate polymers.

🎯 Exam Tip: Know the alternative names and classifications for major biomolecules like carbohydrates, lipids, proteins, and nucleic acids.

 

Question 3. Omnis - cellula e - cellula was given by ........... (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer: (a) Virchow
In simple words: The famous idea "Omnis cellula e cellula," which means "all cells arise from pre-existing cells," was proposed by Rudolf Virchow. This principle is a cornerstone of modern cell theory.

🎯 Exam Tip: Important biological principles and their proponents are frequently tested; remember Rudolf Virchow for his contribution to cell theory.

 

Question 4. Nitrocellulose is used in making
(a) cellophane
(b) drapers
(c) explosives
(d) pain balms
Answer: (c) explosives
In simple words: Nitrocellulose is a special material created from cellulose. It is often used to make explosives because of its highly flammable nature.

🎯 Exam Tip: Be aware of the practical applications of various biomolecules and their derivatives, especially in industrial or chemical contexts.

 

Question 5. Genes present in the cytoplasm of eukaryotic cells are found in ................ (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids
Answer: (a) mitochondria and inherited via egg cytoplasm
In simple words: In eukaryotic cells, most genes are in the nucleus, but some are also found in the mitochondria, which are powerhouses of the cell. These mitochondrial genes are typically passed down from the mother through the egg cell's cytoplasm. This is why mitochondrial DNA is useful for tracing maternal lineages.

🎯 Exam Tip: Remember that mitochondria (and chloroplasts in plants) have their own DNA, which is distinct from nuclear DNA and has a unique inheritance pattern.

 

Question 6. Chitin when added with amino acid becomes
(a) myeopolysaccharide
(b) amylopolysaccharide
(c) mucopolysaccharide
(d) peptidopolysaccharide
Answer: (c) mucopolysaccharide
In simple words: Chitin is a type of sugar chain found in insect exoskeletons. When it combines with amino acids, it forms a larger, more complex molecule called a mucopolysaccharide. This combination is important for structural components in various organisms.

🎯 Exam Tip: Understand the composition and classification of complex carbohydrates, especially how they combine with other molecules like amino acids to form important biological structures.

 

Question 7. A quantosome is present in ............... . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer: (b) Chloroplast
In simple words: Quantosomes are very small particles found inside chloroplasts, which are the parts of plant cells that perform photosynthesis. They contain all the necessary components for capturing sunlight and turning it into energy.

🎯 Exam Tip: Associate specific cellular structures with their unique components; quantosomes are characteristic of chloroplasts and their role in photosynthesis.

 

Question 8. Among the following one is not a non-polar solvent
(a) benzene
(b) sulphuric acid
(c) ether
(d) chloroform
Answer: (b) sulphuric acid
In simple words: Non-polar solvents are liquids that do not have charged ends and usually dissolve non-polar substances. Benzene, ether, and chloroform are examples of non-polar solvents. However, sulphuric acid is a very strong polar acid, meaning it has charged ends and can dissolve polar substances.

🎯 Exam Tip: Understand the difference between polar and non-polar solvents, as this concept is fundamental to understanding solubility and chemical interactions in biology and chemistry.

 

Question 9. One of the given below is a complex found in the cell membrane of animal cell
(a) cholesterol
(b) myelin
(c) proline
(d) lecithin
Answer: (a) cholesterol
In simple words: Cholesterol is a type of fat molecule that is an important part of animal cell membranes. It helps to keep the membrane fluid and strong, preventing it from being too rigid or too flexible. This molecule is vital for the cell's structure and function.

🎯 Exam Tip: Remember cholesterol as a key component of animal cell membranes, playing a crucial role in membrane fluidity and stability.

 

Question 11. Principle information molecules of the cell are known as
(a) Nucleus
(b) DNA
(c) RNA
(d) Nucleic acids
Answer: (d) Nucleic acids
In simple words: Nucleic acids, which include DNA and RNA, are the molecules that carry genetic information in cells. They store, transmit, and express genetic instructions, guiding all cellular activities.

🎯 Exam Tip: Understand that while DNA and RNA are specific types, "nucleic acids" is the broader category for the informational molecules of life.

 

Question 12.
(I) Cellulose – A most abundant organic compound
(II) Morphine – Pain relieving alkaloid
(III) Aldose – reducing sugar & Ketose
(IV) Glycogen – mucopolysaccharide
Answer: (IV) Glycogen – mucopolysaccharide
In simple words: The question asks to identify the incorrect match. Glycogen is a polysaccharide, but it is primarily a storage polysaccharide, not a mucopolysaccharide. Mucopolysaccharides are more complex sugars often found in connective tissues.

🎯 Exam Tip: Pay close attention to classifications and specific definitions of biomolecules to identify incorrect statements in matching questions.

 

Question 13. Identify the given structure.
(a) Amino acid
(b) Fatty acid
(c) Nucleotide
(d) Monosaccharide
Answer: (a) Amino acid
In simple words: The diagram shows a central carbon atom connected to an amino group (N-H), a carboxyl group (C=O-OH), a hydrogen atom, and a variable side chain (R group). This specific arrangement is the basic structure of an amino acid, which are the building blocks of proteins.

🎯 Exam Tip: Be able to visually identify the core structures of basic biomolecules like amino acids, fatty acids, sugars, and nucleotides.

 

Question 14. Lactose is a disaccharide of
(a) Glucose – Glucose
(b) Fructose – Fructose
(c) Glucose – Galactose
(d) Fructose – Galactose
Answer: (c) Glucose – Galactose
In simple words: Lactose is a type of sugar called a disaccharide, meaning it is made of two smaller sugar units joined together. These two units are glucose and galactose. This combination is commonly found in milk and dairy products.

🎯 Exam Tip: Remember the monosaccharide components of common disaccharides (e.g., sucrose = glucose + fructose; maltose = glucose + glucose; lactose = glucose + galactose).

 

Question 15. Number of fatty acids in triglyceride is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: A triglyceride is a type of fat molecule that serves as a major way to store energy in our bodies. It is made up of one glycerol molecule attached to three fatty acid molecules.

🎯 Exam Tip: Know the basic structure of lipids, specifically that a triglyceride consists of a glycerol backbone and three fatty acid chains.

 

Question 16. Heparin the anti-coagulant is got from
(a) D – glucuronic acid
(b) Polymer of fructose
(c) Mucopolysaccharide from red algae
(d) Glycosaminoglycan
Answer: (d) Glycosaminoglycan
In simple words: Heparin is a natural substance in the body that helps prevent blood from clotting. It is classified as a glycosaminoglycan, which is a type of complex carbohydrate with specific sugar units. This unique structure helps it act as an anticoagulant.

🎯 Exam Tip: Understand the role of complex carbohydrates and their derivatives in biological functions, such as heparin's function as an anticoagulant.

 

Question 17. The pH at which Zwitterion is formed is known as
(a) Iso ionic balance
(b) Isoelectric potential
(c) Isoelectric point
(d) Iso ionic point
Answer: (c) Isoelectric point
In simple words: A zwitterion is a molecule, like an amino acid, that has both a positive and a negative charge at different parts, but the overall molecule is neutral. The specific pH level where this electrically neutral zwitterion form is dominant is called its isoelectric point. This point is crucial for how proteins behave in solutions.

🎯 Exam Tip: Distinguish between isoelectric point (pI), where net charge is zero, and isoionic point, which relates to the pH of a solution containing only the pure substance.

 

Question 18. Aspartate and Glutamate are amino acids of
(a) Negatively charged 'R' groups
(b) Positively charged 'R' groups
(c) Non-polar aliphatic 'W groups
(d) Non-polar aromatic 'R' groups
Answer: (a) Negatively charged 'R' groups
In simple words: Aspartate and Glutamate are two amino acids that are known for having negatively charged side chains, also called 'R' groups. This negative charge comes from a carboxyl group in their side chain that loses a proton at normal physiological pH. These amino acids are often involved in reactions that require an acidic environment.

🎯 Exam Tip: Categorize amino acids based on their 'R' group properties (polar, non-polar, charged, aromatic) as this determines their role in protein structure and function.

 

Question 19. The test for protein is
(a) iodine test
(b) Biuret test
(c) Benedict's test
(d) Hydrolysis test
Answer: (b) Biuret test
In simple words: The Biuret test is a common way to check if a protein is present. It uses a special chemical that changes color when it finds protein.

🎯 Exam Tip: Remember that the Biuret test is specific for peptide bonds, which are the links that hold amino acids together in a protein chain.

 

Question 20. The competitive inhibitor is ................for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer: (a) malonate
In simple words: Malonate is a chemical that stops succinic dehydrogenase from working properly. It competes with the enzyme's natural partner.

🎯 Exam Tip: Competitive inhibitors often have a similar shape to the natural substrate, allowing them to bind to the enzyme's active site and block the reaction.

 

Question 21. Formation of new chemical bonds using ATP as a source of energy
a. Lyase
b. Hydrolase
c. Telomerase
d. Ligase
Answer: (d) Ligase
In simple words: Ligases are like tiny molecular 'gluers'. They help create new chemical bonds, using energy from ATP to join things together.

🎯 Exam Tip: The key characteristic of a ligase is its ability to catalyze the formation of a bond coupled with the hydrolysis of a high-energy phosphate bond (like ATP).

 

Question 22. Uridylic acid is an
a. Dinucleotide
b. Nucleoside
c. Nucleotide
d. Ribo nucleotide
Answer: (d) Ribonucleotide
In simple words: Uridylic acid is a basic building block of RNA, which is a type of genetic material. It is called a ribonucleotide because it has a ribose sugar, a phosphate group, and a uracil base.

🎯 Exam Tip: Remember that nucleotides are made of three parts: a sugar, a phosphate group, and a nitrogenous base. The 'ribo' part indicates the sugar is ribose, found in RNA.

 

Question 23. Phosphate forming linkage with sugar is known as
a. diester linkage
b. peptide linkage
c. phosphodiester linkage
d. Ionic linkage
Answer: (c) phosphodiester linkage
In simple words: The special connection between a phosphate group and a sugar in DNA or RNA is called a phosphodiester linkage. It's a strong bond that holds the backbone of these molecules together.

🎯 Exam Tip: This linkage is crucial for the structure of nucleic acids (DNA and RNA), forming the sugar-phosphate backbone.

 

Question 24. ................ is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer: (b) Ribozyme
In simple words: Ribozymes are special RNA molecules that can act like enzymes, helping chemical reactions happen. They show that not only proteins can be catalysts.

🎯 Exam Tip: While most enzymes are proteins, ribozymes are a key example of RNA molecules performing enzymatic functions, highlighting RNA's diverse roles.

 

Question 25. A class of lipid that serves as a major component of the cell membrane is
a. triglyceride
b. glycerol
c. phospho lipid
d. lipoprotein
Answer: (c) phospholipid
In simple words: Phospholipids are like tiny bricks that make up the walls of cells. They form the main part of the cell membrane, helping to control what goes in and out.

🎯 Exam Tip: Phospholipids have a hydrophilic (water-loving) head and hydrophobic (water-fearing) tails, which allows them to form the bilayer structure of cell membranes.

 

Question 26. One molecule of sucrose on hydrolysis give
a. 2 molecules of glucose
b. 1 molecule glucose & 1 molecule fructose
c. 2 molecules of glucose & 1 molecule of fructose
d. 2 molecules of fructose
Answer: (b) 1 molecule glucose & 1 molecule fructose
In simple words: When sucrose, which is common table sugar, is broken down by water, it splits into two simpler sugars: one molecule of glucose and one molecule of fructose.

🎯 Exam Tip: Remember that sucrose is a disaccharide, meaning it's formed from two monosaccharide units. Hydrolysis breaks these units apart.

 

Question 27. In fibrous proteins polypeptide chain are held together by
a. Vander Waals forces
b. disulphide linkage
c. electrostatic forces
d. hydrogen bonds
Answer: (a) Vander Waals forces
In simple words: Fibrous proteins are like strong ropes made of many tiny threads. These threads, or polypeptide chains, are held together by weak forces called Van der Waals forces, making the protein strong and stable.

🎯 Exam Tip: While various forces contribute to protein stability, Van der Waals forces, along with hydrogen bonds, play a significant role in maintaining the structure of fibrous proteins.

 

Question 28. According to Chargaff's rule, the hydrogen bonding between Adenine and
(a) 2
(b) 3
(c) 4
(d) Nil
Answer: (a) 2
In simple words: Chargaff's rule explains how DNA bases pair up. Adenine always connects to Thymine using two hydrogen bonds, like a specific type of sticky glue.

🎯 Exam Tip: Always remember the specific base pairing rules in DNA: Adenine (A) always pairs with Thymine (T) via two hydrogen bonds, and Guanine (G) always pairs with Cytosine (C) via three hydrogen bonds.

 

Question 29. Which polymer is stored in liver
a. Amylose
b. Amylo pectin
c. Cellulose
d. Glycogen
Answer: (d) Glycogen
In simple words: Glycogen is a complex sugar that acts as an energy store in our bodies. The liver holds a lot of this glycogen, releasing sugar when needed for energy.

🎯 Exam Tip: Glycogen is often called "animal starch" because it is the primary storage form of glucose in animals and fungi, similar to how starch stores glucose in plants.

 

Question 30. The bond that is not needed for protein formation is
a. Hydrogen bond
b. Peptide bond
c. Ionic bond
d. glucosidic bond
Answer: (d) glucosidic bond
In simple words: Proteins are built using peptide bonds, helped by hydrogen and ionic bonds. Glucosidic bonds, however, are used for sugars, not proteins.

🎯 Exam Tip: Distinguish between the types of bonds: peptide bonds form proteins, glycosidic bonds form carbohydrates, and phosphodiester bonds form nucleic acids.

 

Question 31. A complete turn of the helix comprises
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer: (b) 3.4 nm
In simple words: In the famous twisted ladder shape of DNA, one full spiral of the ladder measures 3.4 nanometers long. This measurement is key to how DNA is structured.

🎯 Exam Tip: Remember the key dimensions of the B-DNA helix: a diameter of 2 nm, 10 base pairs per turn, and a pitch (length of one complete turn) of 3.4 nm.

 

Question 32. The acid is also known as vitamin C
a. Aspartic acid
b. Tartaric acid
c. Ascorbic acid
d. Adipic acid
Answer: (c) Ascorbic acid
In simple words: Ascorbic acid is simply another name for Vitamin C. This vitamin is important for our body to stay healthy and fight off sickness.

🎯 Exam Tip: Vitamin C (Ascorbic acid) is a water-soluble vitamin known for its role in immune function, collagen synthesis, and antioxidant properties.

 

Question 33. Which is the left-handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer: (c) Z – DNA
In simple words: While most DNA is right-handed (like a right-handed screw), there's a special type called Z-DNA that twists to the left. This difference in twisting affects how it works.

🎯 Exam Tip: Z-DNA is unique among DNA forms due to its left-handed helical structure, which is influenced by specific nucleotide sequences and ionic conditions.

II. Choose the wrong answer.

 

Question 1.
(a) Hydrolase – Amylase
(b) Oxiaoreductase – Dehydrogenase
(c) Transferase – Transaminase
(d) Isomerase – Hexokinase
Answer: (d) Isomerase – Hexokinase
In simple words: Hexokinase is an enzyme that adds a phosphate group, so it's a transferase, not an isomerase. Isomerases only rearrange parts within a molecule.

🎯 Exam Tip: When identifying enzyme classes, focus on the type of reaction they catalyze: hydrolases use water to break bonds, oxidoreductases transfer electrons, transferases move groups, and isomerases rearrange atoms within a molecule.

 

Question 2. This has nothing to do with the structure of

a. Cytosine
b. Pyrimidine
c. Adenine
d. Thyamine
Answer: (c) Adenine
In simple words: The picture shows a chemical structure known as pyrimidine. Cytosine, pyrimidine itself, and thymine are all pyrimidines. Adenine, however, is a purine, which has a different structure, so it doesn't fit with this image.

🎯 Exam Tip: Be able to visually distinguish between purine (double-ring structure, like Adenine and Guanine) and pyrimidine (single-ring structure, like Cytosine, Thymine, and Uracil) bases.

 

Question 3. Choose the right answer
a) Amylose – linear unbranched polymer of with 20% starch
b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure
c) Inulin – Polymer of galactose
d) Amino acid – Here a basic structure of carbon linked to a basic amino group
Answer: (a) Amylose – linear unbranched polymer of with 20% starch
In simple words: Amylose is a part of starch that forms a straight chain, without many branches. It makes up about 20% of the starch we find in plants.

🎯 Exam Tip: Understand the differences between amylose and amylopectin in starch: amylose is linear and unbranched, while amylopectin is highly branched, impacting their properties and digestion.

III. Match The Following And Find The Correct Answer.

 

Question 1.
(I) Morphine
(II) Concanavalin A
(III) Vinblastin
(IV) Anthocyanin

A. Hectins
B. Drug
C. Pigment
D. Toxin

Answer: (a) B-A-D-C
In simple words: Morphine is a drug. Concanavalin A is a type of lectin (a protein). Vinblastin is a toxin often used in medicines. Anthocyanin gives plants their color, so it's a pigment.

🎯 Exam Tip: When matching, categorize each item (e.g., substance, function, class) first, then find the best fit. Knowing common examples helps a lot.

 

Question 2.
(I) Lactose
(II) Rhamnose
(III) Stachyose
(IV) Verbascose

A. Penta saccharide
B. Tetra saccharide
C. Disaccharide
D. Tri saccharide

Answer: (b) C-D-B-A
In simple words: Lactose is a disaccharide (two sugar units). Rhamnose is a monosaccharide, but in the context of these options, D (Trisaccharide) is given as a match in the answer, implying a categorization for complex sugars. Stachyose is a tetrasaccharide (four sugar units). Verbascose is a pentasaccharide (five sugar units).

🎯 Exam Tip: Knowing the number of sugar units in common oligosaccharides (di-, tri-, tetra-, penta-saccharides) is essential for matching questions like these.

 

Question 3.
(I) Amino acid chain is twisted into coiled configuration called a helix
(II) Protein fold into a globular structure called domains
(III) Linear arrangement of amino acids in a Polypeptide chain
(IV) more than one polypeptide forms a large multiunit multimer

A. Tertiary Protein
B. Quaternary protein
C. Secondary protein
D. Primary Protein

Answer: (d) C-B-A-D
In simple words: The basic chain of amino acids is the primary protein structure. When this chain coils into a helix, it forms a secondary structure. A folded globular protein is tertiary. Multiple protein chains joining together make a quaternary structure.

🎯 Exam Tip: Clearly understand the four levels of protein structure: primary (sequence), secondary (local folding like helices), tertiary (overall 3D shape), and quaternary (multiple polypeptide chains interacting).

 

Question 4.
(I) Amino acid chain is twisted into coiled configuration called a helix
(II) Protein fold into a globular structure called domains
(III) Linear arrangement of amino acids in a Polypeptide chain
(IV) more than one polypeptide forms a large multiunit multimer

A. Tertiary Protein
B. Quaternary protein
C. Secondary protein
D. Primary Protein

Answer: (c) C – A – D-B
In simple words: The simple, straight chain of amino acids is the primary structure. Coils or folds, like a helix, are secondary. The overall three-dimensional shape, often globular, is tertiary. When several of these tertiary units join, it becomes a quaternary structure.

🎯 Exam Tip: This question tests your understanding of the hierarchical organization of protein structures. Each level builds upon the last, adding complexity to the protein's function.

 

Question 5.
(I) Esters formed between long-chain alcohol and saturated fatty acid
(II) Di sulphide bridges formed between sulphur & amino acids
(III) The amino acids are both acidic & basic exoskeleton of insects
(IV) Zwitter is also called dipolar

A. a molecule with two or more saturated fatty acid function group one +ve
B. fluid nature & selective
C. waxy substance coating
D. amophoteric in nature

Answer: (c) C-B-D-A
In simple words: Esters like waxes act as a protective coating. Disulphide bridges affect how flexible a protein can be. Amino acids can act as both acid and base, which is called amphoteric. A zwitterion is a molecule with both positive and negative charges, making it dipolar.

🎯 Exam Tip: Pay close attention to the specific definitions of biochemical terms. A zwitterion is a key concept for amino acids, and amphoteric describes their dual acidic/basic nature.

IV. Find Out The True And False Statements From The Following And Find Out The Right Answer.

 

Question 1.
(I) Esters are formed between long-chain alcohol & saturated fatty acid.
(II) Lecithin is a food additive & dietary supplement.
(III) Lipids in their structure have two hydrophilic ends
(IV) Solid fats are usually unsaturated

Answer: (a) True – True – False – False
In simple words: Esters are indeed made from long-chain alcohols and fatty acids. Lecithin is used as a food additive. Lipids typically have one hydrophilic (water-loving) end and one or more hydrophobic (water-fearing) tails. Solid fats are usually saturated, while unsaturated fats are typically liquid at room temperature.

🎯 Exam Tip: Understand the properties of lipids: their hydrophobic and hydrophilic parts, the difference between saturated and unsaturated fats, and common examples like esters and lecithin.

 

Question 2.
(I) In saturated fatty acids, the hydrocarbon chain is single-bonded
(II) Triglycerides are composed of a single molecule of glycerol bound to 2 fatty acids
(III) Palmitic acid is an example of saturated fatty acid.
(IV) Oleic acid is an example of unsaturated fatty acid.

Answer: (c) True – False – True – True
In simple words: Saturated fatty acids only have single bonds. Triglycerides are actually made of one glycerol and *three* fatty acids, not two. Palmitic acid is a known saturated fatty acid. Oleic acid, on the other hand, has double bonds, making it an unsaturated fatty acid.

🎯 Exam Tip: Accurately recall the definition and structure of saturated and unsaturated fatty acids, and the composition of triglycerides (glycerol + three fatty acids).

V.

 

Question 1. Label the diagram parts correctly by choosing the right option.

Answer: (d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide
In simple words: The diagram shows a nucleotide, the building block of DNA. A is the phosphate group, B is the deoxyribose sugar, and C is the nitrogenous base. D represents the entire nucleotide structure.

🎯 Exam Tip: Memorize the three main components of a nucleotide: a phosphate group, a pentose sugar (deoxyribose in DNA, ribose in RNA), and a nitrogenous base.

 

Question 2.

Answer: (b) P arm – Centromere – Sister chromatids – Q arm
In simple words: The picture shows a chromosome. A is the P arm (short arm), C is the centromere (the constriction in the middle), B indicates the sister chromatids (the duplicated halves), and D is the Q arm (long arm).

🎯 Exam Tip: Accurately identify the key parts of a chromosome: sister chromatids (identical copies), centromere (constriction point), and the p and q arms.

VI. Assertion & Reason – Find Out The Correct Answer.

 

Question 1. Assertion (A): Adhesion refers to the tendency of water molecules to cling together
Reason (R): Because of hydrogen bonding, water molecules interact with one another continuous column of water is raised in xylem vessels.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer: (c) Assertion is true but Reason is wrong
In simple words: Adhesion is when water sticks to other things, like the walls of a glass. Cohesion is when water molecules stick to each other. The reason given describes cohesion, not adhesion. Both cohesion and adhesion are important for water movement in plants.

🎯 Exam Tip: Clearly differentiate between adhesion (water sticking to other surfaces) and cohesion (water molecules sticking to each other), as both are vital for water transport in plants.

 

Question 2. Assertion (A): Glycine is a non-essential amino acid
Reason (R): It must be taken through diet
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer: (c) Assertion is true but Reason is wrong.
In simple words: Glycine is indeed a non-essential amino acid, meaning our body can make it. So, it doesn't need to be gotten from food. The reason for being non-essential is not that it must be in the diet, but the opposite.

🎯 Exam Tip: A non-essential amino acid is one that the body can synthesize, thus it is not strictly required from the diet. An essential amino acid must be obtained from the diet.

 

Question 3. Assertion (A): In the presence of enzyme substance molecules can be attached by the reagent.
Reason (R): Active sites of enzymes hold the substance in a suitable position.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion
Answer: (a) Assertion & Reason correct Reason Explaining Assertion.
In simple words: Enzymes help connect molecules together. They do this by having a special spot called an 'active site' that perfectly holds the molecules in the right place so they can join up. This makes the reaction happen smoothly.

🎯 Exam Tip: The lock-and-key or induced-fit models explain how an enzyme's active site precisely binds substrates, facilitating specific biochemical reactions.

 

Question 4. Assertion (A): Aminoacids behave like salt rather than simple amines or carboxylic acid.
Reason (R): In aqueous solution, the COOH group of amino acid loses a proton and the NH2 group accepts a proton to form zwitterion (salt).
(a) Assertion & Reason correct Reason Explaining Assertion
(c) Assertion is true but Reason is wrong
Answer: (a) Assertion & Reason correct Reason Explaining Assertion.
In simple words: Amino acids have both acid and base parts. In water, the acid part gives away a hydrogen, and the base part takes one. This creates a molecule with both positive and negative charges, like a salt, called a zwitterion. This zwitterion form is why they act differently than simple acids or bases.

🎯 Exam Tip: The formation of a zwitterion, with both a positively charged amino group and a negatively charged carboxyl group, is a fundamental property of amino acids in neutral solutions.

2 Marks

 

Question 1. Define Micronutrients.
Answer: Micronutrients are important substances that plants and animals need only in very small amounts to grow and stay healthy. These small amounts are still vital for many body processes. For example, cobalt, zinc, boron, copper, molybdenum, and manganese are micronutrients that help enzymes work correctly.
In simple words: Micronutrients are special food parts needed in tiny amounts. They are very important for plants and animals to grow and for body processes to work.

🎯 Exam Tip: Remember that "trace amounts" or "small quantities" are the defining characteristic of micronutrients, differentiating them from macronutrients which are needed in larger amounts.

 

Question 2. Write down the properties of Water.
Answer: Water has many unique qualities that are important for life.

• It has both adhesion and cohesion properties, meaning it sticks to other surfaces and to itself.
• Water has a high latent heat of vaporization, so it takes a lot of energy for it to turn into vapor.
• It also has a high melting and boiling point compared to other similar molecules.
• Water is known as a universal solvent because it can dissolve many different substances.
• It has a specific heat capacity, meaning it can absorb a lot of heat without a large temperature change. These properties help moderate Earth's climate.

In simple words: Water sticks to itself and other things. It needs a lot of heat to boil or melt. It can dissolve many things and holds heat well.

🎯 Exam Tip: Focus on the properties that stem from hydrogen bonding (cohesion, adhesion, high specific heat, high heat of vaporization) as these are crucial for water's role in biological systems.

 

Question 3. Differentiate between Primary and Secondary Metabolites.
Answer:

In simple words: Primary metabolites are crucial for a plant or animal to live and grow, like the energy it makes. Secondary metabolites are not directly needed for survival but help in other ways, like protecting against enemies.

🎯 Exam Tip: The key difference is their necessity for survival: primary metabolites are essential for life, while secondary metabolites serve specialized, non-essential (but often beneficial) roles.

 

Question 4. Define Polymerisation.
Answer: Polymerization is a chemical process where small, repeating units, called monomers, link together to form long chains known as polymers. This process creates large molecules from smaller ones. For example, starch is a polynucleotide, formed by many glucose units linking together.
In simple words: Polymerization is when many small parts join up to make one very long chain, like building a big Lego structure from many small bricks.

🎯 Exam Tip: Clearly define "monomer" and "polymer" when explaining polymerization to show full understanding.

 

Question 5. Distinguish between Glycogen and Cellulose.
Answer:

In simple words: Glycogen is how animals store sugar for energy, and cellulose is what makes plants strong and stiff, like in wood or cotton. They are both made of sugar but are used for different jobs.

🎯 Exam Tip: When distinguishing between biological molecules, focus on their primary function, typical location, and key structural differences like the type of linkage or branching.

 

Question 6. Distinguish between Dinucleotide & Polynucleotide.
Answer:

In simple words: A dinucleotide is just two small building blocks (nucleotides) joined together, but a polynucleotide is a very long chain made of many, many such blocks. DNA and RNA are examples of polynucleotides.

🎯 Exam Tip: Focus on the "di-" (two) vs. "poly-" (many) prefixes to remember the key difference in the number of nucleotide units.

 

Question 7. Differentiate between Nucleoside & Nucleotide.
Answer:

In simple words: A nucleoside is a base and a sugar. A nucleotide is the same base and sugar, but with an extra phosphate group attached. That phosphate makes nucleotides the real building blocks of DNA.

🎯 Exam Tip: Remember the "P" in Nucleotide stands for Phosphate, which is the extra component compared to a nucleoside.

 

Question 8. State Chargaff's Law.
Answer: Chargaff's Law, proposed in 1949, describes important rules about the composition of DNA. These rules are:
1. The amount of adenine (A) is always equal to the amount of thymine (T), and the amount of guanine (G) is always equal to the amount of cytosine (C). This is because A pairs with T, and G pairs with C.
2. There is a double bond between adenine and thymine, and a triple bond between guanine and cytosine.
3. The total number of purines (A + G) is equal to the total number of pyrimidines (T + C). This maintains a balanced structure in the DNA helix.
4. However, the ratio of (A + T) to (G + C) can vary among different species, meaning A:T is not necessarily equal to G:C across all organisms.
In simple words: Chargaff's Law says that in DNA, A always matches T, and G always matches C. So, you'll find the same amount of A as T, and G as C.

🎯 Exam Tip: The core of Chargaff's Law is \( A=T \) and \( G=C \), and \( A+G=T+C \). Make sure to state these equalities clearly.

 

Question 9. Differentiate Between DNA & RNA.
Answer:

In simple words: DNA is like the main instruction book for life and has two strands, while RNA is like a helper that reads those instructions and makes proteins, usually having only one strand.

🎯 Exam Tip: Remember the two key differences: DNA has deoxyribose sugar and thymine, while RNA has ribose sugar and uracil. Also, DNA is usually double-stranded, and RNA is single-stranded.

 

Question 10. Distinguish between Cation, Zwitterion & Anion.
Answer: Amino acids can exist in different ionic forms depending on the pH of their environment. These forms are cation, zwitterion, and anion, each with a distinct charge:

1. Cation: At very low pH (acidic conditions), the amino acid has a net positive charge. This happens because the amino group \( (\text{-NH}_2) \) picks up a proton and becomes \( (\text{-NH}_3^+) \), and the carboxyl group \( (\text{-COOH}) \) remains protonated. 2. Zwitterion: At a neutral pH, which is specific for each amino acid (called the isoelectric point), the amino acid exists as a zwitterion. In this form, the amino group is positively charged \( (\text{-NH}_3^+) \) and the carboxyl group is negatively charged \( (\text{-COO}^-) \). The overall molecule has a net charge of zero. This unique property allows amino acids to act as buffers. 3. Anion: At high pH (basic conditions), the amino acid has a net negative charge. Here, the carboxyl group \( (\text{-COOH}) \) loses its proton to become \( (\text{-COO}^-) \), and the amino group \( (\text{-NH}_2) \) remains deprotonated.
In simple words: An amino acid can change its electrical charge depending on how acidic or basic its surroundings are. It can be positive (cation), have no overall charge (zwitterion), or be negative (anion).

🎯 Exam Tip: Remember that a zwitterion has both positive and negative charges internally, but a net zero charge overall, occurring at its isoelectric point.

 

Question 11. How will you test reducing sugar?
Answer: The presence of reducing sugars like glucose in a food sample can be identified using Benedict's test. Reducing sugars have an aldehyde or ketone group that can donate electrons. Here's how the test works:
1. Take a small amount of the food sample (e.g., glucose solution, which is an aldehyde) in a test tube.
2. Add an alkaline solution of copper(II) sulfate, also known as Benedict's reagent, to the test tube. Benedict's reagent is a clear blue solution.
3. Heat the mixture gently.
4. If reducing sugar is present, the aldehyde or ketone group will reduce the \( \text{Cu}^{2+} \) ions (which are blue) in the Benedict's reagent to \( \text{Cu}^+ \) ions. This forms a brick-red precipitate of copper(I) oxide.
5. In this reaction, the aldehyde or ketone group itself gets oxidized to a carboxyl group \( (\text{-COOH}) \). The intensity of the brick-red color and the amount of precipitate directly relate to the concentration of the reducing sugar present. If no reducing sugar is present, the solution remains blue. Sucrose is a non-reducing sugar and will not cause this color change. This test is a fundamental method for detecting simple sugars.
In simple words: To test for reducing sugar, you mix your sample with a blue liquid called Benedict's reagent and heat it. If the liquid turns brick-red, it means reducing sugar (like glucose) is there.

🎯 Exam Tip: Clearly state the color change (blue to brick-red precipitate) and identify Benedict's reagent as the key chemical for the test.

 

Question 12. Draw the structure of a basic amino acid.
Answer: A basic amino acid contains an R-group with an extra amino group that can accept a proton, making it basic. Here is the general structure of a basic amino acid:

This general structure shows the central carbon atom (alpha carbon) bonded to an amino group (\( \text{H-N-H} \)), a carboxyl group (\( \text{-C(=O)OH} \)), a hydrogen atom, and a variable side chain (R-group). The R-group determines the specific properties of the amino acid. In a basic amino acid, the R-group itself contains an additional amino group or other nitrogen-containing functional groups, making it capable of picking up a proton and thus contributing to a net positive charge under certain pH conditions.
In simple words: An amino acid is a small building block with a central carbon, an amino group, a carboxyl group, and a special 'R' part. The 'R' part makes each amino acid different, and in a basic one, this 'R' part has an extra amino group that can hold a positive charge.

🎯 Exam Tip: When drawing amino acid structures, always ensure the central carbon (alpha carbon) has four different groups attached: an amino group, a carboxyl group, a hydrogen atom, and the unique R-group.

 

Question 1. Classify Polysaccharides.
Answer: Polysaccharides are complex carbohydrates made up of more than ten monosaccharide units linked together. They are primarily classified into two main types:
1. Homopolysaccharides: These are polysaccharides made up of only one type of monosaccharide unit repeated many times. Examples include:

• Glycogen (storage in animals)
• Starch (storage in plants)
• Cellulose (structural in plants)
• Inulin (fructose polymer)

2. Heteropolysaccharides: These are polysaccharides composed of two or more different types of monosaccharide units, or derivatives of monosaccharides. Examples include:

• Peptidoglycan (found in bacterial cell walls)
• Hyaluronic acid (in connective tissues)
• Chondroitin (in cartilage)
• Keratan sulphate (in cartilage and cornea)
• Agar (from red algae, used in labs)
• Heparin (anticoagulant)

These classifications help us understand their diverse roles, from energy storage and structural support to cell recognition and immune function.
In simple words: Polysaccharides are long chains of sugar units. If all the sugar units are the same, it's a homopolysaccharide. If the sugar units are different or have other parts, it's a heteropolysaccharide.

🎯 Exam Tip: Remember that "homo" means "same" and "hetero" means "different" when classifying polysaccharides by their monomer composition. Provide at least two clear examples for each type.

 

Question 2. Why do we call Glucose and Fructose Isomers -Discuss.
Answer: Glucose and Fructose are called isomers because they share the exact same molecular formula, which is \( \text{C}_6\text{H}_{12}\text{O}_6 \), but they have different structural formulas and arrangements of atoms. This means they are made of the same building blocks but are put together differently. Here are their distinct structural representations:

Glucose is an aldose, meaning it contains an aldehyde functional group, typically at the end of its carbon chain. Fructose, on the other hand, is a ketose, possessing a ketone functional group, usually in the middle of its carbon chain. This difference in the placement of the carbonyl group leads to different chemical properties and metabolic pathways, even though they share the same atomic count. This structural variation is what defines them as isomers.
In simple words: Glucose and fructose are isomers because they both have the same number of carbon, hydrogen, and oxygen atoms (\( \text{C}_6\text{H}_{12}\text{O}_6 \)), but their atoms are arranged differently, giving them different shapes and properties.

🎯 Exam Tip: When discussing isomers, always mention the identical molecular formula but different structural formula. Pointing out glucose as an aldose and fructose as a ketose is a key distinguishing factor.

 

Question 3. Explain the formation of Disaccharide Lactose.
Answer: Lactose, commonly known as milk sugar, is a disaccharide formed by the chemical joining of two simpler sugar units: \( \beta \)-D-galactose and \( \alpha \)-D-glucose. This linkage occurs through a glycosidic bond, specifically a \( \beta \)-1,4 glycosidic linkage. The formation involves a dehydration reaction (condensation), where a water molecule is removed when the two monosaccharides join.

The hydroxyl group from the \( \text{C}_1 \) carbon of galactose and the hydroxyl group from the \( \text{C}_4 \) carbon of glucose react to form the glycosidic bond, releasing a molecule of water. This bond is strong and holds the two sugar units together, making lactose a common energy source found in milk. Lactose needs to be broken down by the enzyme lactase into its constituent monosaccharides for proper digestion and absorption.
In simple words: Lactose is made when one molecule of galactose sugar and one molecule of glucose sugar join together, and a water molecule is removed in the process.

🎯 Exam Tip: Specify the two monosaccharides (galactose and glucose) and the type of bond (glycosidic linkage, \( \beta \)-1,4) formed during lactose synthesis. Mentioning dehydration (condensation) reaction is also crucial.

 

Question 4. Draw the structure of Fatty acid.
Answer: A fatty acid is a long chain of hydrocarbons with a carboxyl group at one end. This structure makes one end polar (the carboxyl group) and the long hydrocarbon tail non-polar. The length and saturation of the hydrocarbon chain determine the fatty acid's properties. Here is a general representation of a fatty acid:

This structure represents a saturated fatty acid with a long chain of methylene groups (\( \text{-CH}_2- \)) and a terminal methyl group (\( \text{-CH}_3 \)) at the other end. The carboxyl group (\( \text{-COOH} \)) is the characteristic functional group that gives fatty acids their acidic properties and ability to form esters.
In simple words: A fatty acid is like a long stick of carbon and hydrogen atoms, with a special acidic part (COOH) at one end. This long stick is what makes fats and oils.

🎯 Exam Tip: When drawing a fatty acid, ensure you clearly show the carboxyl group \( \text{-COOH} \) at one end and a long, straight hydrocarbon chain attached to it.

 

Question 5. Distinguish between Waxes & Steroids.
Answer:

In simple words: Waxes are long, protective coatings that repel water, like on leaves or feathers. Steroids are special ring-shaped molecules that are important parts of cell membranes and act as hormones in the body.

🎯 Exam Tip: Highlight the structural difference (long ester chain for waxes vs. four-ring skeleton for steroids) and their distinct functions (protection/waterproofing vs. structural/hormonal roles).

 

Question 6. Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.
Answer: Amino acids are the basic building blocks of proteins, and their diverse structures are primarily due to their unique side chains (R-groups). Glycine, Alanine, and Valine are simple amino acids classified by their non-polar aliphatic R groups. These R-groups contribute to how proteins fold and interact.

The core structure (amino group, carboxyl group, and alpha carbon) is common to all amino acids, but the R-group (side chain) differentiates them. Glycine has the simplest R-group, just a hydrogen atom. Alanine has a methyl group, while Valine has a branched isopropyl group. These differences are crucial for protein folding and function.
In simple words: Glycine, Alanine, and Valine are three simple types of amino acids. They all have the same basic amino acid parts, but their small "R" side chains are different: Glycine has just a hydrogen, Alanine has a single \( \text{CH}_3 \), and Valine has a branched \( \text{CH(CH}_3\text{)}_2 \) group.

🎯 Exam Tip: When drawing amino acid structures, always correctly position the amino group, carboxyl group, and hydrogen on the central carbon, then focus on the unique R-group for each specific amino acid.

 

Question 7. Distinguish between Macronutrients & Micronutrients.
Answer: Macronutrients and micronutrients are both essential for plant growth and development, but they differ in the quantities required by plants:
1. Macronutrients:

• These are nutrients required by plants in larger quantities for optimal growth and various physiological processes.
• They are crucial for building plant structures, energy metabolism, and maintaining osmotic pressure.
• Examples include Potassium and Calcium, along with Nitrogen, Phosphorus, Magnesium, and Sulfur.

2. Micronutrients:

• These are nutrients required by plants in very small, or trace, amounts.
• Even though needed in tiny quantities, they are vital for enzyme activity and proper functioning of metabolic pathways.
• Examples include Zinc and Boron, along with Iron, Manganese, Copper, Molybdenum, and Chlorine.

Both types of nutrients are important, and a deficiency in either can severely impact plant health and yield. Regular soil testing helps determine if plants are getting enough of each nutrient.
In simple words: Macronutrients are nutrients plants need a lot of, like calcium. Micronutrients are nutrients plants need only tiny amounts of, like zinc, but they are still very important.

🎯 Exam Tip: The primary distinction is the *quantity* required. Provide at least two clear examples for each category to score full marks.

 

Question 8. Define Activation energy?
Answer: Activation energy is a fundamental concept in chemical kinetics, particularly in biological systems:
1. The minimum quantity of energy that reactant molecules must possess in order to undergo a specified chemical reaction is known as activation energy. It's the energy barrier that must be overcome for a reaction to occur.
2. Biological catalysts, such as enzymes, significantly reduce this activation energy. By lowering the energy barrier, enzymes help the reaction proceed much faster without being consumed themselves.
3. Consequently, the rate of a chemical reaction increases if its activation energy decreases, as more reactant molecules will have enough energy to react at any given temperature. This is crucial for life processes, allowing complex reactions to happen quickly at body temperature.
In simple words: Activation energy is the smallest amount of energy needed to start a chemical reaction. Enzymes help by lowering this energy, making reactions happen easily and quickly.

🎯 Exam Tip: Define activation energy as the "minimum energy" required. Emphasize that catalysts (like enzymes) *lower* this energy, which *increases* reaction rates.

 

Question 9. Distinguish between Primary metabolite & Secondary metabolite.
Answer: Primary and secondary metabolites are both organic compounds produced by living organisms, but they differ significantly in their roles and necessity for survival:
1. Primary metabolites:

• These are organic compounds directly involved in the normal growth, development, and reproduction of an organism. They are essential for fundamental metabolic processes.
• Examples include molecules like amino acids, nucleotides, simple sugars, and fatty acids, which are integral to processes such as photosynthesis, respiration, and protein/lipid synthesis. Lipase, a protein, is a primary metabolite crucial for fat digestion.

2. Secondary metabolites:

• These are organic compounds that are not directly involved in the normal growth, development, or reproduction of an organism. They often play roles in defense, signaling, or adaptation to the environment.
• They do not show a direct function in the growth and development of organisms, but instead offer advantages like protection from predators or attraction of pollinators.
• Examples include ricin (a toxin for defense) and gums (for protection).

While primary metabolites are universally found in all organisms, secondary metabolites are often unique to specific species or groups, reflecting specialized adaptations.
In simple words: Primary metabolites are chemicals vital for an organism's basic life functions, like growing and making energy. Secondary metabolites are chemicals that are not directly needed for basic life but help the organism in other ways, like defense or attracting partners.

🎯 Exam Tip: The key difference is "direct involvement in essential life processes." Primary metabolites are *directly* essential for growth, while secondary metabolites provide *indirect* benefits like defense or signaling.

 

Question 10. Give examples for Secondary metabolites.
Answer: Secondary metabolites are organic compounds produced by organisms that are not directly involved in their primary metabolic processes but often have ecological roles, such as defense, signaling, or attraction. Some common examples include:
1. Pigments: Substances that give color, such as Carotenoids (found in carrots, responsible for orange color) and Anthocyanins (responsible for red, purple, and blue colors in flowers and fruits).
2. Alkaloids: Nitrogen-containing organic compounds that often have pharmacological effects. Examples include Morphine (a painkiller) and Codeine (a cough suppressant).
3. Essential oils: Volatile aromatic compounds extracted from plants, known for their scent and often used in perfumes or traditional medicine. Lemongrass oil and Rose oil are examples.
4. Lectins: Proteins that bind to carbohydrates, such as Concanavalin A, often involved in cell-cell recognition.
5. Drugs: Various plant-derived compounds used in medicine, like Vinblastine (an anti-cancer drug) and Curcumin (from turmeric, with anti-inflammatory properties).
6. Polymeric substances: Large molecules like Rubber (for elasticity), Gums (for protection and sealing), and Cellulose (though sometimes classified as primary, it can also play secondary roles in specific contexts).
These diverse compounds highlight the chemical complexity and adaptability of living systems.
In simple words: Secondary metabolites are special chemicals made by living things for jobs like making colors (pigments), making medicines (alkaloids), giving smells (essential oils), or helping with defense.

🎯 Exam Tip: For examples of secondary metabolites, try to recall compounds with distinct ecological or pharmaceutical uses, as these are often their primary roles.

 

Question 11. Draw the various structures of Protein.
Answer: Proteins exhibit complex three-dimensional structures, which are crucial for their diverse functions. These structures are categorized into four hierarchical levels: primary, secondary, tertiary, and quaternary. Each level describes a different aspect of the protein's overall shape.

1. Primary Structure: This is the simplest level, representing the linear sequence of amino acids in the polypeptide chain. It's like the order of letters in a word, and it determines all higher levels of structure. 2. Secondary Structure: This refers to localized folding patterns within the polypeptide chain, stabilized by hydrogen bonds between the amino and carboxyl groups of the polypeptide backbone. The two most common secondary structures are the \( \alpha \)-helix (a spiral shape) and the \( \beta \)-pleated sheet (a folded, zigzag pattern). 3. Tertiary Structure: This is the overall three-dimensional shape of a single polypeptide chain, formed by interactions between the R-groups (side chains) of the amino acids. These interactions include hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions, creating a unique and functional globular or fibrous shape. 4. Quaternary Structure: This level applies to proteins composed of more than one polypeptide chain (subunits). It describes how these multiple polypeptide subunits arrange themselves and interact to form a larger, functional protein complex. Hemoglobin, composed of four subunits, is a classic example.
In simple words: Proteins have four main structures: primary (just the order of amino acids), secondary (simple folds like spirals or zigzags), tertiary (the full 3D shape of one chain), and quaternary (how multiple chains fit together). Each level helps the protein do its job.

🎯 Exam Tip: Clearly define each level of protein structure, emphasizing the type of bonding or interactions involved at each stage (e.g., peptide bonds for primary, hydrogen bonds for secondary, R-group interactions for tertiary, and subunit interactions for quaternary).

 

Question 12. Draw the structure of Purine – Adenine & Guanine Nucleotides.
Answer: Purines are a type of nitrogenous base characterized by a double-ring structure, which includes a six-membered ring fused to a five-membered ring. Adenine and Guanine are the two main purine bases found in DNA and RNA. When these bases are attached to a sugar (ribose or deoxyribose) and a phosphate group, they form purine nucleotides. Here are their structures:

Both Adenine and Guanine nucleotides feature the purine double-ring system. Adenine is distinguished by an amino group \( (\text{-NH}_2) \) at position C6 of its six-membered ring, while Guanine has a carbonyl group \( (\text{=O}) \) at position C6 and an amino group \( (\text{-NH}_2) \) at position C2. The sugar and phosphate parts are identical in both, forming the backbone of nucleic acids. These specific chemical differences dictate their base-pairing properties and overall roles in genetic information.
In simple words: Adenine and Guanine are two main purine nucleotides, which means they both have a special double-ring shape. They are part of DNA and RNA. Adenine has an \( \text{NH}_2 \) group, and Guanine has an oxygen atom and an \( \text{NH}_2 \) group in different spots. Both are joined to a sugar and a phosphate.

🎯 Exam Tip: Focus on the common purine double-ring structure, and then highlight the specific functional groups (amino, carbonyl) that differentiate adenine from guanine at particular carbon positions.

 

Question 13. Why do some people have curly hair?
Answer: Curly hair is primarily determined by the shape of the hair follicle and the distribution of disulfide bonds within the hair protein, keratin. Human hair is predominantly made of keratin, a fibrous protein. The more disulfide (sulfur-sulfur) bonds form between the cysteine amino acids within the keratin structure, and the more irregularly these bonds are spaced along the hair shaft, the more the protein chains bend and fold. This increased bending and folding cause the hair strands to curl. A genetic predisposition dictates the quantity and placement of these disulfide bonds, making some individuals naturally have curly hair. Additionally, the cross-sectional shape of the hair follicle (oval follicles produce curlier hair, round follicles produce straighter hair) also influences hair texture.
In simple words: Curly hair happens because of how the protein in our hair is shaped. Small bonds called disulfide bonds make the hair strands bend and twist, causing curls. The more these bonds are spread out and formed, the curlier the hair becomes.

🎯 Exam Tip: The key factors for curly hair are the presence and arrangement of disulfide bonds (sulfur atoms) in keratin, and the cross-sectional shape of the hair follicle.

 

Question 14. Deoxyribose (C5H10O4) is not a carbohydrate – Discuss.
Answer: Deoxyribose, with the chemical formula \( \text{C}_5\text{H}_{10}\text{O}_4 \), is not strictly classified as a carbohydrate, even though it is a sugar component of DNA. Carbohydrates are typically defined by the general formula \( \text{C}_n(\text{H}_2\text{O})_m \), where the ratio of hydrogen to oxygen is usually 2:1, similar to water. While \( \text{C}_5\text{H}_{10}\text{O}_4 \) can be written as \( \text{C}_5(\text{H}_2\text{O})_4 \), it has one oxygen atom less than what would be expected for a typical pentose sugar (like ribose, \( \text{C}_5\text{H}_{10}\text{O}_5 \), which fits \( \text{C}_5(\text{H}_2\text{O})_5 \)). This "deoxygenated" nature (lacking one oxygen compared to ribose) means it doesn't perfectly fit the traditional "hydrate of carbon" definition where the carbon and oxygen atoms are in a 1:1 ratio. This structural difference is critical, as it makes DNA more stable than RNA, given the missing hydroxyl group. Thus, while it functions as a sugar, its specific molecular structure places it outside the strict classic definition of a carbohydrate due to the absence of that one oxygen atom.
In simple words: Deoxyribose has the formula \( \text{C}_5\text{H}_{10}\text{O}_4 \). Even though it's a sugar, it's not strictly a carbohydrate because it has one less oxygen atom than a typical sugar like ribose, which makes its hydrogen-to-oxygen ratio not exactly like water.

🎯 Exam Tip: The core reason deoxyribose is not a "true" carbohydrate by classical definition is its missing oxygen atom compared to ribose, leading to a different \( \text{C}_n(\text{H}_2\text{O})_m \) ratio (specifically, \( m \neq n \) when \( n=5 \)).

5 Marks Questions

 

Question 1. How will you identify the presence of glucose in a given food sample?
Answer: The presence of glucose, a common reducing sugar, in a food sample can be identified using Benedict's test. This test relies on the reducing properties of glucose's aldehyde group. Here's a detailed procedure and explanation:
1. Preparation: Obtain a liquid sample of the food to be tested. If the food is solid, it should first be crushed and dissolved in distilled water.
2. Adding Reagent: Transfer about 2 ml of the food sample into a clean test tube. Then, add an equal volume (around 2 ml) of Benedict's reagent to the test tube. Benedict's reagent is a light blue solution containing copper(II) sulfate \( (\text{CuSO}_4) \), sodium citrate, and sodium carbonate.
3. Heating: Gently heat the test tube in a boiling water bath for 5-10 minutes. The mixture should be observed carefully during this period.
4. Observation and Interpretation:
* No Change (Stays Blue): If no reducing sugars are present (e.g., if only sucrose, a non-reducing sugar, is in the sample), the solution will remain blue. * Color Change: If glucose or other reducing sugars are present, a color change will occur. The blue Benedict's reagent will turn green, then yellow, then orange, and finally a brick-red precipitate will form. The progression of color (and the amount of precipitate) indicates the concentration of reducing sugar: green for low concentration, yellow for moderate, orange for high, and brick-red for very high concentration.
Chemical Principle: The aldehyde group of glucose reduces the blue \( \text{Cu}^{2+} \) ions in Benedict's reagent to red \( \text{Cu}^+ \) ions, forming insoluble copper(I) oxide \( (\text{Cu}_2\text{O}) \), which is the brick-red precipitate. In this redox reaction, glucose itself is oxidized to gluconic acid. This chemical transformation is very sensitive to the amount of reducing sugar. The more reducing sugar present, the greater the reduction of \( \text{Cu}^{2+} \) ions, leading to a more intense brick-red color and heavier precipitate.
In simple words: To find glucose, you use Benedict's test. Mix the food sample with blue Benedict's liquid and heat it. If glucose is there, the blue liquid will change colors (green, yellow, orange) and finally turn brick-red with a grainy powder. The deeper the red, the more glucose there is.

🎯 Exam Tip: Remember the color sequence of Benedict's test (blue \( \rightarrow \) green \( \rightarrow \) yellow \( \rightarrow \) orange \( \rightarrow \) brick-red precipitate) and that it detects *reducing* sugars. Mentioning the chemical reduction of \( \text{Cu}^{2+} \) to \( \text{Cu}^+ \) is key.

 

Question 2. What is protein denaturation.
Answer: Protein denaturation is a process where a protein loses its specific three-dimensional structure (secondary, tertiary, and sometimes quaternary structures) without breaking its primary structure (the sequence of amino acids). This loss of structure typically leads to a loss of the protein's biological function. Denaturation happens due to external stress or compounds that disrupt the non-covalent interactions and disulfide bonds essential for the protein's folded shape.
Causes of Denaturation:
1. Heat: High temperatures cause atoms within the protein to vibrate violently. This excessive kinetic energy overcomes the weaker hydrogen and ionic bonds, causing the protein to unfold. For example, cooking an egg denatures the egg white proteins.
2. Extreme pH: Drastic changes in pH (very acidic or very basic conditions) alter the charges on the amino acid side chains. This disrupts the ionic bonds and hydrogen bonds, leading to changes in the protein's shape.
3. Chemical Agents: Substances like strong acids, bases, heavy metal salts, alcohols, and detergents can disrupt the interchain bonds and hydrophobic interactions within the protein. Soaps and detergents, for instance, can interfere with hydrophobic interactions, causing proteins to lose their folded structure. Disinfectants also work by denaturing proteins in microorganisms.
When a protein denatures, it often becomes elongated, disorganised strands or aggregates, making it non-functional. In many cases, denaturation is irreversible, meaning the protein cannot refold back into its active form.
In simple words: Protein denaturation is when a protein loses its original folded shape, usually because of heat, strong acids, or other chemicals. When it loses its shape, it can't do its job anymore, like an egg white changing when you cook it.

🎯 Exam Tip: Define denaturation as the loss of 3D structure (not primary structure) and explain that it results in loss of function. Give at least two common causes (heat, pH) and briefly explain how they work.

 

Question 3. Tabulate other sugar compounds
Answer: Here is a tabulation of various sugar compounds, primarily polysaccharides, detailing their structures and functions:

In simple words: This table shows different types of complex sugars (polysaccharides) and their roles. For example, inulin helps check kidney function, hyaluronic acid makes joints flexible, and heparin stops blood from clotting.

🎯 Exam Tip: For each sugar compound, remember its basic structural classification (homo/heteropolysaccharide, polymer of what) and one primary function or application.

 

Question 4. Classify enzyme reactions:
Answer: Enzyme reactions can be classified in several ways, reflecting their diverse roles in metabolism and their location within a cell:
I. Based on Metabolic Role (Anabolic vs. Catabolic):
1. Anabolic reactions (building up): These are synthesis reactions where enzymes help combine smaller molecules to form larger, more complex molecules. These reactions typically require energy.
* Examples: Synthesis of proteins from amino acids, or the creation of polysaccharides from simple sugars.
2. Catabolic reactions (breaking down): These are breakdown reactions where enzymes help split large, complex molecules into smaller, simpler ones. These reactions often release energy.
* Examples: Digestion of complex food particles into simpler nutrients, or cellular respiration where sugar (glucose) is broken down to produce energy (glycolysis).
II. Based on Location (Extracellular vs. Intracellular):
1. Extracellular enzymes: These are enzymes secreted outside the cell where they perform their functions. They work externally to break down substances before they are absorbed by the cell.
* Examples: Digestive enzymes like amylase and pepsin, which break down food in the digestive tract.
2. Intracellular enzymes: These enzymes function inside the cell. They are typically involved in cellular metabolic pathways within the cytoplasm or organelles.
* Examples: Enzymes involved in glycolysis (like hexokinase) or the Krebs cycle, and even hormones like insulin, which regulates glucose metabolism inside cells.
This classification helps us understand where and how enzymes exert their catalytic power to maintain life processes.
In simple words: Enzymes help in two main kinds of reactions: "building up" (anabolic), where small things make big things, and "breaking down" (catabolic), where big things become small. Enzymes also work either outside the cell (extracellular) or inside the cell (intracellular) to do their jobs.

🎯 Exam Tip: Clearly define anabolic (synthesis, consumes energy) and catabolic (breakdown, releases energy), and extracellular (outside cell) vs. intracellular (inside cell), providing a relevant example for each classification.

 

Question 5. Explain the three types of Co-Factors.
Answer: Co-factors are non-protein chemical compounds or metallic ions that are required by enzymes for their activity as catalysts. They can be thought of as "helper molecules" that assist enzymes in carrying out specific biochemical reactions. Without their appropriate co-factors, many enzymes would be inactive. There are three main types of co-factors:
1. Inorganic ions: These are metal ions that bind to the enzyme, helping it to form a proper active site or to stabilize the enzyme-substrate complex. For example, salivary amylase, an enzyme that breaks down starch, requires chloride ions \( (\text{Cl}^{++}) \) to increase its activity. Other examples include \( \text{Mg}^{2+} \), \( \text{Fe}^{2+} \), and \( \text{Zn}^{2+} \).
2. Prosthetic groups: These are organic molecules that are tightly (covalently or non-covalently) bound to the enzyme. They are an integral part of the enzyme's structure. For instance, Vitamin \( \text{B}_2 \) (Riboflavin) is a component of Flavin Adenine Dinucleotide (FAD), which is a prosthetic group in the Krebs cycle. The heme group in hemoglobin is also a prosthetic group.
3. Co-enzymes: These are organic molecules that are loosely bound to the enzyme during the reaction, often carrying chemical groups or electrons between enzymes. Unlike prosthetic groups, co-enzymes can detach from the enzyme after the reaction and participate in multiple reactions. Examples include \( \text{NAD}^+ \) (Nicotinamide Adenine Dinucleotide), \( \text{NADP}^+ \) (Nicotinamide Adenine Dinucleotide Phosphate), and Co-enzyme A. These are often derived from vitamins.
Co-factors are crucial for enzyme function, enabling the vast array of biochemical reactions necessary for life.
In simple words: Co-factors are like small helpers that enzymes need to work. They can be metal ions (inorganic ions), organic molecules tightly attached to the enzyme (prosthetic groups), or organic molecules that temporarily join the enzyme to help with the reaction (co-enzymes).

🎯 Exam Tip: Distinguish between the three types based on their chemical nature (inorganic vs. organic) and how tightly they bind to the enzyme (tightly bound prosthetic groups vs. loosely bound co-enzymes).

 

Question 6. Tabulate the uses of enzymes
Answer: Enzymes have a wide range of applications in various industries due to their high specificity and efficiency as biological catalysts. Here are some common uses of enzymes:

In simple words: Enzymes are used in many ways, like in washing powders to remove stains (protease), to make sweet syrup (glucose isomerase), or to remove lactose from milk (lactase). They help make many products better.

🎯 Exam Tip: For each enzyme, remember its source (e.g., bacteria, fungi) and one clear industrial or domestic application where its catalytic function is utilized.

 

Question 7. Enumerate the properties of Enzyme.
Answer:

• Enzymes are special proteins that are shaped like tiny balls (globular).
• They act as catalysts, which means they speed up chemical reactions, and they are effective even in small amounts.
• Enzymes do not change themselves during the reaction process.
• Each enzyme is very specific and usually works only on certain reactions or molecules.
• They have a special part called an "active site" where the chemical reaction takes place.
• Enzymes help reactions start more easily by lowering the amount of energy needed (activation energy).

In simple words: Enzymes are tiny protein helpers that speed up specific reactions without changing themselves, using a special spot called an active site. They make reactions happen with less effort.

🎯 Exam Tip: Focus on keywords like "catalyst," "specific," "active site," and "lower activation energy" to describe enzyme properties correctly.

 

Question 8. Explain lock and key mechanism of enzymes.
Answer: The lock and key mechanism describes how an enzyme works with its specific substrate. Imagine an enzyme as a lock with a unique shape, and the molecule it acts upon (called the substrate) as a key. This model suggests that the substrate fits perfectly into a special pocket on the enzyme, known as the "active site," much like a specific key fits into a specific lock. Once the substrate (key) binds to the active site (lock), the enzyme facilitates the chemical change, turning the substrate into a new product. This precise fit ensures that enzymes are highly specific in their actions.
In simple words: An enzyme acts like a lock, and the molecule it works on (substrate) acts like a key. They fit together perfectly at a special spot called the active site, allowing the enzyme to do its job.

🎯 Exam Tip: Clearly define the enzyme as the "lock" and the substrate as the "key" and explain how their specific fit leads to a reaction.

 

Question 9. What are the various types of inhibitors of enzymes.
Answer: Inhibitors are substances that reduce or stop the activity of enzymes in cells. There are generally three main types of enzyme inhibitors:

• I. Competitive Inhibitors: These substances have a shape similar to the enzyme's natural substrate, allowing them to compete and bind to the enzyme's active site. By occupying the active site, they prevent the actual substrate from binding and reacting. For example, oxygen or carbon dioxide can competitively inhibit the RUBISCO enzyme, and malonate inhibits succinic dehydrogenase.
• II. Non-Competitive Inhibitors: Unlike competitive inhibitors, these molecules bind to a different site on the enzyme, not the active site. This binding changes the enzyme's overall shape, which then affects the active site and prevents the substrate from binding or reacting effectively. For instance, amino acid Alanine can inhibit pyruvate kinase.
• III. Non-reversible/Irreversible Inhibitors: These inhibitors form strong, permanent bonds with the enzyme, often at or near the active site. This binding permanently alters or destroys the enzyme's catalytic ability. Examples include cyanide ions inhibiting cytochrome oxidase, and nerve gas sarin blocking neurotransmitters.

In simple words: Enzyme inhibitors are substances that slow down or stop enzymes. Competitive inhibitors look like the enzyme's target and block its active spot. Non-competitive inhibitors change the enzyme's shape from another spot. Irreversible inhibitors stick permanently and ruin the enzyme's function.

🎯 Exam Tip: Remember to explain how each type of inhibitor works, focusing on whether it binds to the active site or another part of the enzyme, and whether the effect is temporary or permanent.

 

Question 10. Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.
Answer: The two terms are closely related, with negative feedback (end product) inhibition being a specific type of allosteric inhibition:

• Feedback Allosteric Inhibition: This happens when an inhibitor molecule binds to a site on the enzyme that is different from the active site, called the allosteric site. This binding changes the enzyme's shape, which then affects its active site and reduces or stops its activity. This effect is often reversible. For example, glucose-6-phosphate can allosterically inhibit hexokinase.
• Negative Feedback (End Product) Inhibition: This is a common regulatory mechanism in metabolic pathways and is a specific form of allosteric inhibition. In this process, the final product of a series of enzymatic reactions accumulates and then binds to an enzyme early in the pathway (at an allosteric site). This binding inhibits the enzyme's activity, slowing down or stopping the entire pathway, which in turn reduces the production of the end product. Once the end product concentration decreases, the inhibition is lifted, and the pathway resumes.

In simple words: Allosteric inhibition means a molecule changes an enzyme's shape by binding to a special spot, not the main active part. Negative feedback is when the final product of a series of reactions does this to an earlier enzyme, telling the process to slow down.

🎯 Exam Tip: Clearly define both types of inhibition, highlighting where the inhibitor binds (active site vs. allosteric site) and how end-product inhibition regulates metabolic pathways.

 

Question 11. Tabulate other sugar compounds.
Answer:

In simple words: This table lists different types of sugar compounds, what they are made of, and what jobs they do in nature or for us. For example, Inulin is a fructose sugar that passes through the body, while Hyaluronic acid makes body parts flexible.

🎯 Exam Tip: Organize the information clearly in a table with columns for the sugar compound, its structure/composition, and its main biological function.