Topic: Enzymes; Subtopic: Structure and Function of Enzymes
Keyword Definitions:
Enzyme: Biological catalyst that speeds up biochemical reactions without being consumed.
Cofactor: Non-protein chemical compound required for enzyme activity.
Coenzyme: Organic molecule serving as a transient carrier of specific atoms or functional groups in enzyme-catalyzed reactions.
Apoenzyme: The protein part of an enzyme without its cofactor, inactive until combined with it.
Prosthetic group: Tightly bound non-protein component essential for enzyme activity.
Lead Question (2025):
The protein portion of an enzyme is called:
(1) Cofactor
(2) Coenzyme
(3) Apoenzyme
(4) Prosthetic group
Explanation: The protein part of an enzyme, devoid of its non-protein cofactor, is known as the **apoenzyme**. It is catalytically inactive alone but becomes active upon binding with its cofactor, forming a holoenzyme. Apoenzymes determine enzyme specificity and are crucial for the catalytic process. Hence, the correct answer is (3) Apoenzyme.
1. Which of the following correctly describes the complete enzyme including its cofactor?
(1) Apoenzyme
(2) Holoenzyme
(3) Coenzyme
(4) Isoenzyme
Explanation: When an apoenzyme combines with its cofactor, it forms a catalytically active enzyme known as a **holoenzyme**. The apoenzyme provides specificity, while the cofactor enables catalysis. Together, they function efficiently in metabolic pathways. Therefore, the correct answer is (2) Holoenzyme.
2. The non-protein organic part of an enzyme which is loosely attached is called:
(1) Prosthetic group
(2) Coenzyme
(3) Apoenzyme
(4) Metal ion
Explanation: A **coenzyme** is an organic, non-protein molecule that temporarily associates with an enzyme to assist catalysis. Unlike prosthetic groups, coenzymes bind loosely and may be released after the reaction. Vitamins often serve as precursors of coenzymes. Hence, the correct answer is (2) Coenzyme.
3. Which among the following enzymes requires a metal ion as a cofactor?
(1) Carbonic anhydrase
(2) Urease
(3) Catalase
(4) Sucrase
Explanation: **Carbonic anhydrase** requires Zn²⁺ as a cofactor for its catalytic action. The zinc ion helps in the reversible hydration of carbon dioxide, forming carbonic acid. Enzymes depending on metal ions are called metalloenzymes. Therefore, the correct answer is (1) Carbonic anhydrase.
4. An enzyme and its substrate form a temporary complex called:
(1) Holoenzyme
(2) Enzyme-substrate complex
(3) Apoenzyme
(4) Coenzyme complex
Explanation: During catalysis, the substrate binds to the enzyme’s active site forming an **enzyme-substrate complex**. This interaction lowers activation energy, allowing the reaction to proceed faster. Once the product is formed, it is released and the enzyme remains unchanged. Hence, the correct answer is (2) Enzyme-substrate complex.
5. Which of the following statements about enzymes is correct?
(1) Enzymes increase activation energy
(2) Enzymes are consumed in reactions
(3) Enzymes lower activation energy
(4) Enzymes change equilibrium of reactions
Explanation: **Enzymes lower the activation energy** required for chemical reactions by stabilizing transition states. This increases reaction rate without altering the equilibrium or being consumed. They function under mild physiological conditions. Hence, the correct answer is (3) Enzymes lower activation energy.
6. Which of the following correctly represents the inactive form of an enzyme?
(1) Zymogen
(2) Holoenzyme
(3) Apoenzyme
(4) Isoenzyme
Explanation: Many enzymes are synthesized in an inactive precursor form called **zymogen** or proenzyme. They require biochemical modification, such as cleavage of peptide bonds, to become active. This mechanism prevents unwanted enzymatic activity within cells. Hence, the correct answer is (1) Zymogen.
Assertion–Reason Type Question
7. Assertion (A): Apoenzyme alone is catalytically inactive.
Reason (R): Apoenzyme requires a cofactor to form the active holoenzyme.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, R is false.
(4) A is false, R is true.
Explanation: The **apoenzyme** lacks its cofactor and cannot catalyze reactions until combined with it to form a holoenzyme. Therefore, both statements are true, and R correctly explains A. Hence, the correct answer is (1) Both A and R are true, and R is the correct explanation of A.
Matching Type Question
8. Match the following enzymes with their cofactors:
List I (Enzyme) – List II (Cofactor)
A. Carbonic anhydrase 1. Cu²⁺
B. Urease 2. Zn²⁺
C. Cytochrome oxidase 3. Ni²⁺
D. Catalase 4. Fe³⁺
(1) A–2, B–3, C–1, D–4
(2) A–1, B–4, C–2, D–3
(3) A–3, B–2, C–4, D–1
(4) A–4, B–1, C–3, D–2
Explanation: Carbonic anhydrase uses Zn²⁺, urease uses Ni²⁺, cytochrome oxidase uses Cu²⁺, and catalase requires Fe³⁺ for activity. These cofactors are essential for catalytic reactions. Therefore, the correct matching is A–2, B–3, C–1, D–4, corresponding to option (1).
Fill in the Blanks Question
9. The complete, catalytically active enzyme consisting of an apoenzyme and its cofactor is called __________.
(1) Isoenzyme
(2) Holoenzyme
(3) Zymogen
(4) Allosteric enzyme
Explanation: The combination of an apoenzyme and its cofactor forms a **holoenzyme**, the active form capable of catalyzing reactions. The cofactor provides essential chemical groups or ions, while the apoenzyme offers substrate specificity. Hence, the correct answer is (2) Holoenzyme.
Choose the Correct Statements (Statement I & II)
10. Statement I: Apoenzyme combines with cofactor to form holoenzyme.
Statement II: Cofactor is always a metal ion.
(1) Both statements are true.
(2) Both statements are false.
(3) Statement I true, Statement II false.
(4) Statement I false, Statement II true.
Explanation: **Statement I is true** because an apoenzyme forms a holoenzyme upon binding with its cofactor. **Statement II is false**, as cofactors may be metal ions or organic coenzymes. Hence, the correct answer is (3) Statement I true, Statement II false.
Topic: Structure and Function of Nucleotides; Subtopic: Components of Nucleic Acids
Keyword Definitions:
• Nucleotide: Basic structural unit of nucleic acids, composed of a nitrogen base, a sugar, and a phosphate group.
• Nucleoside: Combination of a nitrogen base and a sugar, without a phosphate group.
• Amino acid: Organic compound containing amino and carboxyl groups, serving as the building blocks of proteins.
• Nitrogen base: Purine or pyrimidine component that pairs to form the genetic code in DNA or RNA.
• Adenine: A purine nitrogen base found in both DNA and RNA.
Lead Question - 2025
List I with List II:
List I List II
A. Adenosine I. Nitrogen base
B. Adenylic acid II. Nucleotide
C. Adenine III. Nucleoside
D. Alanine IV. Amino acid
Choose the option with all correct matches
(1) A-III, B-IV, C-II, D-I
(2) A-III, B-II, C-IV, D-I
(3) A-III, B-II, C-I, D-IV
(4) A-II, B-III, C-I, D-IV
Explanation: Adenosine is a nucleoside formed by adenine and ribose. Adenylic acid (AMP) is a nucleotide containing adenine, ribose, and phosphate. Adenine is a nitrogen base, and alanine is an amino acid found in proteins. Therefore, the correct matches are A–III, B–II, C–I, and D–IV. Hence, the correct answer is (3). Nucleotides form nucleic acids, while amino acids form proteins, both essential for life processes.
1. Which of the following components form a nucleotide?
(1) Sugar + Base
(2) Sugar + Base + Phosphate
(3) Base + Phosphate
(4) Sugar + Phosphate
Explanation: A nucleotide consists of a nitrogen base, a pentose sugar, and one or more phosphate groups. This structure makes it the monomeric unit of nucleic acids, participating in genetic coding and energy transfer reactions like ATP. The correct answer is (2). Each nucleotide is linked by phosphodiester bonds forming DNA/RNA chains.
2. Which of the following bases is not found in RNA?
(1) Uracil
(2) Adenine
(3) Thymine
(4) Cytosine
Explanation: RNA contains adenine, uracil, cytosine, and guanine. Thymine is absent in RNA and replaced by uracil. This distinction helps differentiate DNA and RNA molecules during transcription and translation processes. Therefore, the correct answer is (3). DNA uses thymine, while RNA uses uracil for complementary base pairing.
3. Which type of bond links nucleotides in a nucleic acid chain?
(1) Hydrogen bond
(2) Ionic bond
(3) Phosphodiester bond
(4) Peptide bond
Explanation: Adjacent nucleotides in a nucleic acid chain are connected through phosphodiester bonds between the 3′ hydroxyl group of one sugar and the 5′ phosphate of the next. These covalent bonds provide backbone stability to DNA and RNA. The correct answer is (3). Hydrogen bonds occur between complementary bases, not between nucleotides.
4. The sugar in DNA is
(1) Ribose
(2) Deoxyribose
(3) Glucose
(4) Fructose
Explanation: DNA contains deoxyribose sugar, lacking one oxygen atom at the 2′ carbon compared to ribose in RNA. This difference gives DNA greater stability for long-term genetic storage. The correct answer is (2). Deoxyribose combines with phosphate and bases to form the double helical structure of DNA, discovered by Watson and Crick.
5. In DNA, adenine pairs with
(1) Cytosine
(2) Thymine
(3) Guanine
(4) Uracil
Explanation: In DNA, adenine pairs specifically with thymine through two hydrogen bonds, following Chargaff’s base pairing rule. This complementary pairing ensures genetic stability and accurate replication. The correct answer is (2). In RNA, adenine pairs with uracil instead of thymine, maintaining the same purine–pyrimidine pairing ratio.
6. The base sequence in DNA that codes for a specific protein is called a
(1) Codon
(2) Anticodon
(3) Gene
(4) Chromosome
Explanation: A gene is a specific DNA segment containing information for the synthesis of a protein or RNA molecule. Codons are triplets in mRNA that direct amino acid sequence, while genes represent the full hereditary unit. The correct answer is (3). Genes regulate traits and biological processes through controlled expression.
7. Assertion-Reason Question:
Assertion (A): DNA replication is semiconservative.
Reason (R): In replication, one parental strand acts as a template for the formation of a new complementary strand.
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Explanation: DNA replication is semiconservative because each daughter DNA retains one parental strand and synthesizes a new one. Meselson and Stahl confirmed this experimentally using N15 labeling. Hence, both statements are true and R correctly explains A. The correct answer is (1).
8. Matching Type Question:
Match List I with List II:
List I List II
A. Guanine I. Purine
B. Cytosine II. Pyrimidine
C. Ribose III. Pentose sugar
D. Peptide bond IV. Proteins
(1) A-I, B-II, C-III, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-I, B-III, C-II, D-IV
(4) A-II, B-I, C-III, D-IV
Explanation: Guanine and adenine are purines, while cytosine, thymine, and uracil are pyrimidines. Ribose is a pentose sugar in RNA, and peptide bonds link amino acids in proteins. Thus, the correct matches are A–I, B–II, C–III, and D–IV. The correct answer is (1).
9. Fill in the Blanks Question:
DNA is composed of __________ types of nitrogen bases.
(1) Two
(2) Three
(3) Four
(4) Five
Explanation: DNA contains four types of nitrogen bases: adenine, thymine, guanine, and cytosine. They pair in a specific pattern, A–T and G–C, forming the genetic code. These four bases store all hereditary information. Hence, the correct answer is (3). Their precise arrangement determines gene function and protein synthesis.
10. Choose the Correct Statements Question:
Statement I: DNA contains thymine, while RNA contains uracil.
Statement II: RNA is generally double-stranded in eukaryotic cells.
(1) Both statements are correct
(2) Both statements are incorrect
(3) Statement I is correct and Statement II is incorrect
(4) Statement I is incorrect and Statement II is correct
Explanation: DNA contains thymine as a base, whereas RNA replaces thymine with uracil. Most RNA molecules are single-stranded in eukaryotic cells, unlike the double-stranded DNA. Hence, the correct answer is (3). Structural and base differences distinguish their functions in protein synthesis and heredity.
Topic: Enzymes; Subtopic: Enzyme Classification and Catalysis
Keyword Definitions:
• Enzyme: A biological catalyst that speeds up chemical reactions.
• Substrate (S): The molecule upon which an enzyme acts.
• Transferase: Enzymes that catalyze the transfer of a specific group from one molecule to another.
• Ligase: Enzymes that join two molecules using ATP.
Lead Question - 2025
Name the class of enzyme that usually catalyze the following reaction:
S – G + S′ → S + S′ – G
Where, G is a group other than hydrogen, S is a substrate, and S′ is another substrate.
(1) Hydrolase
(2) Lyase
(3) Transferase
(4) Ligase
Explanation:
Transferases are enzymes that catalyze the transfer of functional groups such as methyl, amino, or phosphate groups between donor and acceptor molecules. In the given reaction, group G is transferred from one substrate (S) to another (S′), which is a hallmark function of transferases. Hydrolases catalyze hydrolysis, lyases remove groups forming double bonds, and ligases form new bonds with ATP. Hence, the correct answer is Transferase. This reaction type is essential in metabolism, biosynthesis, and signal transduction, helping in molecular rearrangement without direct energy input.
1. Which of the following is not a function of enzymes?
(1) Catalyze biochemical reactions
(2) Change equilibrium constant of a reaction
(3) Lower activation energy
(4) Increase rate of reaction
Explanation:
Enzymes increase the rate of biochemical reactions by lowering activation energy but do not alter the equilibrium constant. The equilibrium depends on thermodynamic properties of reactants and products. Enzymes only accelerate reaching equilibrium. Therefore, option (2) is correct. Enzymes are highly specific, reusable, and sensitive to temperature and pH. They form enzyme–substrate complexes and ensure cellular efficiency in metabolic pathways.
2. The enzyme DNA polymerase is classified under which category?
(1) Oxidoreductase
(2) Ligase
(3) Transferase
(4) Hydrolase
Explanation:
DNA polymerase catalyzes the addition of nucleotides to the growing DNA strand using a DNA template, transferring deoxyribonucleotide monophosphate groups. It thus belongs to the transferase category, specifically nucleotidyl transferase. Ligases join DNA fragments, oxidoreductases catalyze oxidation-reduction reactions, and hydrolases catalyze bond cleavage by water. Hence, the correct answer is (3) Transferase. This enzyme plays a vital role in replication and repair processes.
3. Which enzyme catalyzes the joining of two molecules using ATP?
(1) Lyase
(2) Ligase
(3) Hydrolase
(4) Transferase
Explanation:
Ligases catalyze the joining (ligation) of two molecules using energy derived from ATP hydrolysis. They are crucial in DNA replication, repair, and biosynthetic reactions. Lyases break bonds without hydrolysis, hydrolases use water to break bonds, and transferases move functional groups between molecules. Therefore, the correct answer is (2) Ligase. This enzyme helps form stable covalent bonds essential for cellular synthesis.
4. Which of the following statements about enzymes is correct?
(1) Enzymes change the overall ΔG of the reaction
(2) Enzymes increase the activation energy
(3) Enzymes speed up reactions without being consumed
(4) Enzymes permanently alter their structure during reaction
Explanation:
Enzymes accelerate biochemical reactions by lowering activation energy while remaining unchanged after catalysis. They do not alter ΔG (free energy change) or equilibrium. Instead, they stabilize the transition state. Thus, option (3) is correct. Enzymes exhibit specificity due to their active site structure, which precisely fits substrates. After catalysis, the enzyme is regenerated to act repeatedly.
5. Which factor does not affect enzyme activity?
(1) Temperature
(2) pH
(3) Pressure
(4) Light intensity
Explanation:
Enzyme activity is influenced by temperature, pH, substrate concentration, and inhibitors, but not directly by light intensity. Pressure may affect gaseous reactions but has little effect on enzymes in solution. Hence, the correct answer is (4) Light intensity. Enzymes denature at extreme conditions, and maintaining optimal pH and temperature ensures proper conformation and catalytic efficiency for metabolic reactions.
6. Which of the following statements is true for enzyme inhibition?
(1) Competitive inhibitors bind to active site
(2) Non-competitive inhibitors bind to substrate
(3) Competitive inhibitors alter enzyme shape
(4) Non-competitive inhibitors compete with substrate
Explanation:
Competitive inhibitors resemble the substrate and bind to the enzyme’s active site, blocking substrate access. Non-competitive inhibitors bind elsewhere, altering enzyme conformation. Thus, the correct answer is (1) Competitive inhibitors bind to active site. This inhibition can be overcome by increasing substrate concentration, maintaining metabolic control and preventing overproduction of biomolecules in cellular processes.
7. Assertion-Reason Type Question
Assertion (A): Enzymes are specific in action.
Reason (R): Each enzyme acts only on one type of substrate due to complementary active site shape.
(1) Both A and R are true and R explains A
(2) Both A and R are true but R does not explain A
(3) A is true but R is false
(4) A is false but R is true
Explanation:
Enzyme specificity arises because the active site has a unique 3D shape that fits only a specific substrate, forming an enzyme–substrate complex. Hence, both Assertion and Reason are true, and R correctly explains A. Therefore, option (1) is correct. This property maintains regulation and order in biochemical pathways, preventing unwanted reactions in living systems.
8. Matching Type Question
Match the following enzymes with their functions:
A. Oxidoreductase — (i) Transfer of electrons
B. Transferase — (ii) Transfer of functional groups
C. Hydrolase — (iii) Hydrolysis of bonds
D. Lyase — (iv) Addition/removal of groups without hydrolysis
(1) A–i, B–ii, C–iii, D–iv
(2) A–ii, B–i, C–iv, D–iii
(3) A–iii, B–iv, C–i, D–ii
(4) A–iv, B–iii, C–ii, D–i
Explanation:
Enzymes are classified based on reaction type: Oxidoreductases catalyze redox reactions (transfer of electrons), Transferases transfer functional groups, Hydrolases perform bond cleavage using water, and Lyases add or remove groups forming double bonds. Hence, the correct match is (1) A–i, B–ii, C–iii, D–iv. These classes together account for the major catalytic mechanisms in metabolism and biosynthesis.
9. Fill in the Blanks Question
The enzyme that catalyzes conversion of sucrose into glucose and fructose is __________.
(1) Maltase
(2) Sucrase
(3) Lactase
(4) Amylase
Explanation:
Sucrase (invertase) catalyzes the hydrolysis of sucrose into glucose and fructose. This enzyme acts on the α-1,2-glycosidic bond in sucrose. Hence, the correct answer is (2) Sucrase. It is present in intestinal mucosa and essential for sugar digestion. Deficiency leads to intolerance to sucrose-containing foods, causing gastrointestinal disturbances and energy absorption inefficiency.
10. Choose the Correct Statements Question
Statement I: Enzyme activity is measured in International Units (IU).
Statement II: One IU equals the amount of enzyme converting 1 micromole of substrate per minute.
(1) Both statements are correct
(2) Both statements are incorrect
(3) Only Statement I is correct
(4) Only Statement II is correct
Explanation:
Enzyme activity quantifies catalytic efficiency. One International Unit (IU) is defined as the amount of enzyme that catalyzes the conversion of 1 micromole of substrate per minute under defined conditions. Therefore, both statements are correct, making option (1) correct. Measuring enzyme activity helps in enzyme kinetics, diagnostics, and biotechnology applications to assess metabolic health and efficiency.
Topic: Nucleic Acids; Subtopic: RNA World Hypothesis and Evolution of DNA
Keyword Definitions:
• RNA World: The hypothesis that RNA was the first genetic material capable of storing information and catalyzing reactions.
• Genetic Material: Molecules responsible for storing and transmitting hereditary information (DNA or RNA).
• Catalyst: A substance that increases the rate of a biochemical reaction without being consumed.
• DNA: Deoxyribonucleic acid, a stable double-stranded molecule storing genetic information in all living organisms except some viruses.
• Double Helix: The twisted-ladder structure of DNA formed by complementary base pairing.
• Repair Mechanism: Cellular process that corrects errors or damages in DNA to maintain genetic stability.
Lead Question – 2025
Given below are two statements:
Statement – I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement – II: DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both statements I and statement II are correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct but statement II is incorrect
(4) Statement I is incorrect but statement II is correct
Explanation:
Both statements I and II are correct. The RNA world hypothesis suggests that RNA acted as the first genetic material capable of both storing genetic information and catalyzing reactions. However, due to its instability, DNA evolved as a more stable molecule. DNA’s complementary strands and repair systems ensure greater genetic fidelity and evolutionary advantage.
1. Which among the following proves that RNA could have been the first genetic material?
(1) RNA can store information and act as an enzyme
(2) RNA is double-stranded and stable
(3) RNA cannot act as an enzyme
(4) RNA is chemically inert
Explanation: RNA can act both as a genetic material and as a catalyst (ribozyme). This dual function supports the RNA world hypothesis, suggesting that RNA preceded DNA and proteins in evolution. RNA’s catalytic nature enabled early biochemical reactions essential for life formation, even before DNA or enzymes evolved.
2. Which type of RNA acts as a template during protein synthesis?
(1) mRNA
(2) tRNA
(3) rRNA
(4) snRNA
Explanation: Messenger RNA (mRNA) carries genetic information from DNA to the ribosome, where it serves as a template for protein synthesis. The sequence of nucleotides in mRNA determines the sequence of amino acids, ensuring proper translation of genetic information into functional proteins.
3. Which enzyme is responsible for catalyzing DNA synthesis from an RNA template?
(1) DNA polymerase
(2) RNA polymerase
(3) Reverse transcriptase
(4) Ligase
Explanation: Reverse transcriptase synthesizes DNA from an RNA template. It is found in retroviruses such as HIV. This enzyme supports the idea that RNA could have preceded DNA, as it demonstrates that RNA can serve as a template for the evolution of DNA, leading to more stable genetic storage systems.
4. Which of the following gives DNA its greater stability compared to RNA?
(1) Presence of uracil
(2) Presence of thymine and deoxyribose
(3) Presence of ribose sugar
(4) Lack of phosphodiester bonds
Explanation: DNA’s stability arises from the presence of thymine instead of uracil and deoxyribose instead of ribose. Deoxyribose lacks an oxygen atom at the 2′ position, making DNA less reactive and more stable. This chemical property allows DNA to serve as the long-term genetic material in living organisms.
5. Which of the following statements about DNA and RNA is incorrect?
(1) DNA contains thymine, RNA contains uracil
(2) DNA is single-stranded, RNA is double-stranded
(3) Both DNA and RNA contain adenine and guanine
(4) DNA has deoxyribose sugar while RNA has ribose sugar
Explanation: Option (2) is incorrect. DNA is typically double-stranded, while RNA is usually single-stranded. Their difference in sugars and bases makes RNA more reactive but less stable, and DNA more suited for hereditary information storage due to its stable double-helical structure and repair mechanisms.
6. In DNA, adenine pairs with thymine through:
(1) One hydrogen bond
(2) Two hydrogen bonds
(3) Three hydrogen bonds
(4) Phosphodiester bonds
Explanation: In DNA, adenine pairs with thymine through two hydrogen bonds, while cytosine pairs with guanine via three hydrogen bonds. These complementary base pairings ensure genetic fidelity during replication and transcription processes, making DNA a reliable molecule for long-term genetic information storage.
7. Assertion-Reason Type:
Assertion (A): DNA is more stable than RNA.
Reason (R): DNA lacks the 2'-hydroxyl group present in RNA.
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Explanation: Both A and R are true and R correctly explains A. DNA’s deoxyribose lacks a hydroxyl group at the 2′ position, reducing chemical reactivity and enhancing stability. This chemical difference ensures DNA’s resistance to hydrolysis and allows it to serve as a stable genetic material.
8. Matching Type:
Match the following:
A. mRNA 1. Carries amino acids
B. tRNA 2. Template for protein synthesis
C. rRNA 3. Forms ribosomal structure
(1) A–2, B–1, C–3
(2) A–3, B–2, C–1
(3) A–1, B–3, C–2
(4) A–2, B–3, C–1
Explanation: The correct match is A–2, B–1, C–3. mRNA serves as the template for protein synthesis, tRNA brings amino acids to the ribosome, and rRNA forms the structural and catalytic part of ribosomes. Together, they coordinate the translation process of gene expression efficiently.
9. Fill in the Blank:
DNA replication proceeds in the ______ direction.
(1) 3′ → 5′
(2) 5′ → 3′
(3) Random
(4) Both directions equally
Explanation: DNA replication occurs in the 5′ → 3′ direction. DNA polymerase adds nucleotides only to the 3′ end of the growing strand. The leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously as Okazaki fragments, ensuring accurate and semiconservative DNA replication.
10. Choose the Correct Statements (Statement I & II):
Statement I: RNA can catalyze certain biochemical reactions.
Statement II: DNA can act as an enzyme in metabolic reactions.
(1) Both statements are correct
(2) Statement I is correct, Statement II is incorrect
(3) Both statements are incorrect
(4) Statement I is incorrect, Statement II is correct
Explanation: Statement I is correct while Statement II is incorrect. Some RNA molecules (ribozymes) possess catalytic activity, playing roles in RNA splicing and peptide bond formation. DNA, however, is chemically stable and does not exhibit catalytic activity. This further supports RNA’s ancient biological significance.
Subtopic: Reductionist Approach in Biology
Keyword Definitions:
• Reductionist Biology: Study of living systems by breaking them into simpler components.
• Holistic Biology: Approach that studies living organisms as complete systems.
• Physico-chemical approach: Analysis based on physical and chemical principles.
• Organism: An individual life form capable of growth and reproduction.
• Biological organization: Hierarchical structure from molecules to ecosystems.
• Metabolism: Sum of all biochemical reactions in a living organism.
Lead Question - 2025
Which one of the following statements refers to Reductionist Biology?
(1) Physico-chemical approach to study and understand living organisms
(2) Physiological approach to study and understand living organisms
(3) Chemical approach to study and understand living organisms
(4) Behavioural approach to study and understand living organisms
Explanation:
Reductionist Biology explains life by analyzing its simplest physico-chemical and molecular aspects. It focuses on how complex biological phenomena emerge from chemical and physical interactions of molecules and cells. This approach has contributed greatly to molecular biology and biochemistry. Hence, the correct answer is (1) Physico-chemical approach to study and understand living organisms.
1. Which of the following is an example of Reductionist Biology?
(1) Study of DNA replication using enzymes
(2) Observation of animal behavior
(3) Study of ecology of forests
(4) Study of evolution of species
Explanation:
Reductionist Biology involves breaking complex systems into smaller parts for detailed analysis. Studying DNA replication using enzymes represents this molecular-level approach. It focuses on biochemical mechanisms rather than the organism as a whole. Thus, the correct answer is (1) Study of DNA replication using enzymes.
2. Which of the following is a Holistic approach example?
(1) Study of enzyme kinetics
(2) Study of photosynthesis in isolated chloroplasts
(3) Study of an entire ecosystem
(4) Analysis of protein folding
Explanation:
The holistic approach studies living systems in totality, focusing on interactions within the entire system. The study of an ecosystem includes all organisms, abiotic factors, and their relationships, representing a complete picture of life processes. Hence, (3) Study of an entire ecosystem is correct.
3. The holistic approach in biology emphasizes
(1) Interaction of parts within a system
(2) Isolated study of cell components
(3) Molecular study of enzymes
(4) Analysis of atomic structure
Explanation:
The holistic approach stresses understanding the interdependence and integration of biological components within a system. It focuses on how the combined functioning of organs and organisms contributes to life. Hence, the correct answer is (1) Interaction of parts within a system.
4. Who proposed that life can be explained through physico-chemical principles?
(1) Darwin
(2) Mendel
(3) Schwann
(4) Wohler
Explanation:
Friedrich Wohler demonstrated in 1828 that urea could be synthesized from inorganic compounds, proving that organic molecules could arise from physical and chemical reactions. This finding laid the foundation of the reductionist concept in biology. Thus, the correct answer is (4) Wohler.
5. Which of the following best describes the holistic view of biology?
(1) Analysis of molecular components only
(2) Understanding interactions between biological systems
(3) Studying structure of DNA
(4) Measuring enzyme activity
Explanation:
Holistic biology emphasizes that living organisms are more than the sum of their parts. It studies how systems and their components interact to produce emergent properties of life. Therefore, the correct answer is (2) Understanding interactions between biological systems.
6. Which of the following statements is NOT true about Reductionist Biology?
(1) It helps in understanding molecular basis of life
(2) It involves dividing systems into parts for study
(3) It neglects interactions between components
(4) It considers the system as a whole
Explanation:
Reductionist Biology focuses on breaking down living systems into smaller parts to understand their functioning but often overlooks the holistic interactions. It does not consider the system as a whole, making option (4) incorrect about reductionist approach. Hence, the correct answer is (4).
Assertion-Reason Question
7. Assertion (A): Reductionist Biology focuses on studying biological systems at molecular and cellular levels.
Reason (R): The holistic approach ignores molecular studies.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, but R is false.
(4) A is false, but R is true.
Explanation:
Reductionist Biology indeed studies life processes at molecular and cellular levels. However, the holistic approach does not ignore molecular studies but rather integrates them into a system-wide understanding. Hence, A is true, R is false. Correct answer: (3).
Matching Type Question
8. Match the following approaches with their features:
A. Reductionist Biology
B. Holistic Biology
C. Physiology
D. Ecology
List II – Features
(1) Studies whole system interaction
(2) Focus on biochemical pathways
(3) Studies organismal functions
(4) Focus on organism-environment relationship
(a) A–2, B–1, C–3, D–4
(b) A–3, B–2, C–4, D–1
(c) A–4, B–1, C–2, D–3
(d) A–1, B–2, C–3, D–4
Explanation:
Reductionist Biology focuses on biochemical details (2), Holistic Biology on overall system interaction (1), Physiology on functions of organisms (3), and Ecology on environmental relationships (4). Hence, the correct match is (a) A–2, B–1, C–3, D–4.
Fill in the Blanks Question
9. The ________ approach in biology considers the organism as a whole rather than a collection of parts.
(1) Chemical
(2) Reductionist
(3) Holistic
(4) Anatomical
Explanation:
The holistic approach considers the organism as an integrated whole, focusing on the interactions and interdependence of its components rather than individual parts. It provides a broader understanding of life functions. Hence, the correct answer is (3) Holistic.
Choose the Correct Statements Question
10. Statement I: Reductionist Biology simplifies complex biological systems for analysis.
Statement II: Holistic Biology studies parts without considering interrelations.
(1) Both statements are true
(2) Both statements are false
(3) Statement I is true, Statement II is false
(4) Statement I is false, Statement II is true
Explanation:
Reductionist Biology simplifies biological systems into components to understand underlying mechanisms, making Statement I true. Holistic Biology emphasizes interrelations among parts, so Statement II is false. Hence, the correct answer is (3).
Topic: Enzymes; Subtopic: Mechanism of Enzyme Action
Keyword Definitions:
• Enzyme: Biological catalyst that speeds up biochemical reactions without being consumed.
• Active site: Specific region on the enzyme where substrate molecules bind and react.
• Substrate: The reactant molecule upon which the enzyme acts.
• Enzyme-substrate complex: Temporary intermediate formed when enzyme binds to substrate.
• Induced fit model: Explains that substrate binding induces a conformational change in the enzyme to facilitate catalysis.
Lead Question - 2024 (Jhajjhar)
The catalytic cycle of an enzyme action is described as:
A. Enzyme releases products of the reaction and gets free.
B. Substrate induces the enzyme to alter its shape.
C. The substrate binds with the active site of the enzyme.
D. Enzyme-product complex is formed.
1. C → B → A → D
2. C → B → D → A
3. B → C → D → A
4. D → C → A → B
Explanation:
The catalytic cycle begins when the substrate binds to the enzyme’s active site forming the enzyme-substrate complex. This induces structural changes (induced fit model), leading to catalysis and formation of the enzyme-product complex. The enzyme then releases the product and becomes free for reuse. Hence, the correct sequence is C → B → D → A. Enzymes function by lowering activation energy, ensuring rapid biochemical reactions in living cells.
1. The part of the enzyme where the substrate binds is called:
1. Allosteric site
2. Active site
3. Catalytic site
4. Binding site
Explanation:
The active site is the specific region of an enzyme where substrate molecules attach through weak bonds. It has a unique three-dimensional structure complementary to the substrate. The binding leads to the formation of the enzyme-substrate complex, initiating the catalytic reaction. Thus, the correct answer is active site.
2. Which one of the following statements is true about enzymes?
1. They alter the equilibrium of reaction.
2. They lower the activation energy of reaction.
3. They are consumed during reactions.
4. They act at any temperature.
Explanation:
Enzymes lower the activation energy of biochemical reactions without changing the equilibrium point. They speed up reactions by stabilizing the transition state. However, they are not consumed in the process. They function best at optimal pH and temperature. Hence, the correct answer is that enzymes lower activation energy of reactions.
3. Enzymes are made up of:
1. Lipids
2. Proteins
3. Carbohydrates
4. Vitamins
Explanation:
Most enzymes are proteins with a specific three-dimensional conformation essential for their catalytic activity. Some enzymes may also require non-protein cofactors or coenzymes for complete functionality. Protein enzymes can be denatured by heat or pH changes. Hence, the correct answer is proteins.
4. Which of the following models explains enzyme action more accurately?
1. Lock and key model
2. Induced fit model
3. Collision model
4. Random fit model
Explanation:
The induced fit model proposed by Koshland explains that enzyme shape slightly alters when substrate binds to form a perfect fit. This enhances catalytic efficiency. It replaced the older rigid lock and key model, emphasizing enzyme flexibility. Hence, the correct answer is induced fit model.
5. The enzyme that converts starch into maltose is:
1. Maltase
2. Amylase
3. Lactase
4. Sucrase
Explanation:
Amylase catalyzes the hydrolysis of starch into maltose and dextrins. It is secreted by salivary glands and pancreas. It initiates carbohydrate digestion by breaking α-1,4 glycosidic bonds. Hence, the correct answer is amylase, a key digestive enzyme in humans.
6. Which factor does not affect enzyme activity?
1. Temperature
2. pH
3. Substrate concentration
4. Atomic number
Explanation:
Enzyme activity depends on temperature, pH, and substrate concentration, but not on atomic number. Temperature affects enzyme conformation, and extreme pH can denature enzymes. Substrate concentration determines reaction rate until saturation. Hence, the correct answer is atomic number.
Assertion-Reason Type Question
7. Assertion (A): Enzyme action is highly specific.
Reason (R): Enzymes can catalyze reactions of any substrate.
1. Both (A) and (R) are correct, and (R) explains (A).
2. Both (A) and (R) are correct, but (R) does not explain (A).
3. (A) is correct, but (R) is incorrect.
4. (A) is incorrect, but (R) is correct.
Explanation:
Enzyme action is highly specific as each enzyme acts only on a particular substrate due to complementary shape and charge of its active site. Enzymes cannot catalyze reactions of any substrate. Thus, the assertion is correct but the reason is incorrect.
Matching Type Question
8. Match the following:
A. Urease — I. Hydrolysis of starch
B. Amylase — II. Hydrolysis of urea
C. Lipase — III. Breakdown of fats
1. A-II, B-I, C-III
2. A-I, B-II, C-III
3. A-III, B-I, C-II
4. A-II, B-III, C-I
Explanation:
Urease catalyzes the hydrolysis of urea into ammonia and CO₂. Amylase acts on starch to form maltose. Lipase breaks down lipids into fatty acids and glycerol. Hence, the correct match is A-II, B-I, C-III.
Fill in the Blanks Type Question
9. The enzyme which catalyzes the conversion of hydrogen peroxide into water and oxygen is ________.
1. Oxidase
2. Catalase
3. Hydrolase
4. Ligase
Explanation:
Catalase is an enzyme found in most aerobic organisms that decomposes toxic hydrogen peroxide into water and oxygen, preventing oxidative cell damage. It ensures cellular protection from reactive oxygen species. Hence, the correct answer is catalase.
Choose the Correct Statements (Statement I & II)
10. Statement I: Enzyme activity decreases beyond optimum temperature.
Statement II: High temperature denatures the enzyme protein.
1. Both statements are correct.
2. Both statements are incorrect.
3. Statement I is correct, but Statement II is incorrect.
4. Statement I is incorrect, but Statement II is correct.
Explanation:
Both statements are correct. Enzymes show maximum activity at their optimum temperature (usually 37°C in humans). Beyond this, heat denatures the protein structure of the enzyme, rendering it inactive. Therefore, enzyme activity declines at higher temperatures.
Topic: Enzymes and Biological Molecules; Subtopic: Ribozyme, Lecithin, GLUT-4, Vitamins
Keyword Definitions:
Ribozyme: An RNA molecule with catalytic activity capable of acting as an enzyme without proteins.
Lecithin: A phospholipid that plays a vital role in fat metabolism and forms part of the cell membrane.
GLUT-4: A glucose transporter protein found in adipose tissue and skeletal muscles, insulin-dependent.
Vitamin: An organic compound essential in small amounts, often functioning as a coenzyme or precursor.
Coenzyme: A non-protein organic molecule that binds temporarily with enzymes to assist catalysis.
Lipid: Fat-like substances that are insoluble in water and form structural components of membranes.
Enzyme: A biological catalyst that speeds up biochemical reactions without being consumed.
Lead Question - 2024 (Jhajjhar)
Match List-I with List-II:
List-I List-II
A. Ribozyme I. Glucose transport
B. Lecithin II. Non proteinaceous enzyme
C. Glut-4 III. Lipid
D. Vitamins IV. Coenzyme
Choose the correct answer from the options given below:
1. A-IV, B-III, C-I, D-II
2. A-IV, B-III, C-II, D-I
3. A-II, B-I, C-III, D-IV
4. A-II, B-III, C-I, D-IV
Explanation: Ribozyme is a non-protein enzyme (II), Lecithin is a lipid (III), GLUT-4 is a glucose transporter (I), and Vitamins act as coenzymes (IV). Hence, the correct match is A-II, B-III, C-I, D-IV. Ribozymes show catalytic activity, Lecithin supports cell membranes, GLUT-4 helps glucose uptake, and Vitamins function in metabolism.
1. Guessed Question: Which of the following acts as a non-protein enzyme?
1. Immunoglobulin
2. Ribozyme
3. Albumin
4. Hormone
Explanation: Ribozymes are catalytic RNA molecules functioning as enzymes without protein structure. They catalyze RNA splicing and peptide bond formation. This unique property supports the RNA world hypothesis. Hence, Ribozyme acts as a non-protein enzyme.
2. Guessed Question: GLUT-4 transporters are mainly found in:
1. Liver cells
2. Neurons
3. Skeletal muscles and adipose tissue
4. Erythrocytes
Explanation: GLUT-4 is an insulin-dependent glucose transporter present in skeletal muscles and adipose tissue. It facilitates glucose entry during insulin stimulation. This maintains blood glucose homeostasis. Therefore, option 3 is correct.
3. Guessed Question: Lecithin is a phospholipid that is structurally composed of:
1. Glycerol, fatty acids, phosphate, and choline
2. Ribose and phosphate only
3. Protein and carbohydrate
4. Cholesterol only
Explanation: Lecithin is a phospholipid comprising glycerol backbone, fatty acids, phosphate group, and choline. It forms cell membrane bilayers and aids in lipid metabolism. Hence, the correct answer is glycerol, fatty acids, phosphate, and choline.
4. Guessed Question: Vitamins acting as coenzymes participate mainly in:
1. Hormonal balance
2. Enzymatic reactions
3. Protein transport
4. Cell division
Explanation: Vitamins often act as coenzymes or their precursors in metabolic pathways. They assist enzymes in catalyzing biochemical reactions efficiently. For example, vitamin B₁ acts as thiamine pyrophosphate, a coenzyme in carbohydrate metabolism.
5. Guessed Question: Which vitamin acts as a coenzyme for transamination reactions?
1. Vitamin B₁₂
2. Vitamin B₆
3. Vitamin D
4. Vitamin C
Explanation: Vitamin B₆ (pyridoxine) acts as a coenzyme in transamination reactions by forming pyridoxal phosphate. It helps in amino acid metabolism and neurotransmitter synthesis. Therefore, the correct answer is Vitamin B₆.
6. Guessed Question: Identify the biomolecule acting as both structural and metabolic component:
1. Lecithin
2. Collagen
3. Hemoglobin
4. Insulin
Explanation: Lecithin is a phospholipid contributing structurally to the cell membrane and metabolically to lipid transport and emulsification. It stabilizes membranes and aids in fat digestion. Thus, Lecithin has dual functional importance.
7. Assertion-Reason Question:
Assertion (A): Ribozymes are RNA molecules showing catalytic activity.
Reason (R): They contain an active protein site responsible for catalysis.
1. Both A and R are true and R is the correct explanation
2. Both A and R are true but R is not the correct explanation
3. A is true but R is false
4. A is false but R is true
Explanation: Ribozymes are catalytic RNAs that function without proteins; their catalytic sites are RNA-based, not protein-based. Hence, A is true but R is false.
8. Matching Type Question:
Match the following enzymes with cofactors:
A. Carbonic anhydrase – (i) Zn²⁺
B. Hexokinase – (ii) Mg²⁺
C. Catalase – (iii) Fe³⁺
D. DNA polymerase – (iv) Mn²⁺
Options:
1. A-i, B-ii, C-iii, D-iv
2. A-iii, B-i, C-ii, D-iv
3. A-ii, B-iii, C-i, D-iv
4. A-i, B-ii, C-iv, D-iii
Explanation: Correct pairing is A-i, B-ii, C-iii, D-iv. Each enzyme requires specific metal cofactors for function. Carbonic anhydrase uses Zn²⁺, Hexokinase needs Mg²⁺, Catalase has Fe³⁺, and DNA polymerase is activated by Mn²⁺ ions.
9. Fill in the Blanks:
The glucose transporter that is insulin-dependent is _______.
1. GLUT-1
2. GLUT-2
3. GLUT-3
4. GLUT-4
Explanation: GLUT-4 is the insulin-dependent glucose transporter found in skeletal muscles and adipose tissue. It mediates glucose uptake after insulin binding. This mechanism maintains normal blood sugar levels and cellular energy balance.
10. Choose the Correct Statements (Statement I & Statement II):
Statement I: Vitamins act as enzymes for metabolic reactions.
Statement II: Vitamins act as coenzymes, assisting enzyme function.
1. Both statements are true
2. Both statements are false
3. Statement I is true, Statement II is false
4. Statement I is false, Statement II is true
Explanation: Vitamins themselves are not enzymes; they serve as coenzymes or precursors, supporting enzyme activity in metabolism. Therefore, Statement I is false, and Statement II is true.
Topic: Enzymes and Proteins; Subtopic: Types and Functions of Biomolecules
Keyword Definitions:
Collagen: A structural protein found abundantly in connective tissues like tendons and skin.
GLUT-4: A glucose transporter protein that mediates glucose uptake into muscle and fat cells.
Trypsin: A digestive enzyme that breaks down proteins in the small intestine.
RuBisCO: Ribulose-1,5-bisphosphate carboxylase/oxygenase, an enzyme involved in photosynthesis.
Lead Question - 2024 (Jhajjhar)
Match List-I with List-II
List-I List-II
A. Collagen I. Enzyme
B. GLUT-4 II. Most abundant enzyme in biosphere
C. Trypsin III. Most abundant protein in animal world
D. RuBisCO IV. Enables glucose transport into cells
1. A-III, B-I, C-IV, D-II
2. A-II, B-III, C-I, D-IV
3. A-III, B-IV, C-I, D-II
4. A-I, B-III, C-IV, D-II
Explanation: The correct answer is option 3 (A-III, B-IV, C-I, D-II). Collagen is the most abundant protein in animals, providing structural strength. GLUT-4 facilitates glucose entry into cells. Trypsin acts as a digestive enzyme, and RuBisCO is the most abundant enzyme in the biosphere responsible for carbon fixation during photosynthesis. These biomolecules ensure metabolic efficiency and structural integrity across organisms.
1. Which of the following enzymes catalyzes the breakdown of starch into maltose?
1. Maltase
2. Sucrase
3. Amylase
4. Lactase
Explanation: The correct answer is option 3 (Amylase). Amylase, secreted by salivary glands and pancreas, hydrolyzes starch into maltose and dextrins. It is essential for carbohydrate digestion. This enzyme functions optimally at near-neutral pH in the mouth and slightly alkaline conditions in the small intestine, facilitating efficient conversion of polysaccharides to disaccharides.
2. Which of the following bonds is broken during the denaturation of proteins?
1. Peptide bonds
2. Hydrogen bonds
3. Glycosidic bonds
4. Phosphodiester bonds
Explanation: The correct answer is option 2 (Hydrogen bonds). Denaturation involves the disruption of weak hydrogen and ionic interactions without breaking peptide bonds. This leads to loss of the native conformation and biological activity of proteins. Heat, pH changes, or chemicals like urea can induce denaturation, making proteins biologically inactive but chemically intact.
3. Which among the following enzymes is responsible for the conversion of fibrinogen to fibrin during blood clotting?
1. Pepsin
2. Trypsin
3. Thrombin
4. Amylase
Explanation: The correct answer is option 3 (Thrombin). Thrombin converts soluble fibrinogen into insoluble fibrin, forming a blood clot mesh. This conversion is a critical part of the coagulation cascade, preventing excessive bleeding after injury. The enzyme is produced from prothrombin by prothrombinase in the presence of calcium ions and clotting factors.
4. Assertion-Reason Question:
Assertion (A): RuBisCO is the most abundant enzyme in the biosphere.
Reason (R): It catalyzes the carboxylation of RuBP in the Calvin cycle during photosynthesis.
1. Both A and R are true, and R is the correct explanation of A
2. Both A and R are true, but R is not the correct explanation of A
3. A is true, R is false
4. A is false, R is true
Explanation: The correct answer is option 1. RuBisCO catalyzes the carboxylation of ribulose-1,5-bisphosphate (RuBP) to form 3-phosphoglycerate, a key step in photosynthetic carbon fixation. Due to its abundance in all photosynthetic organisms, RuBisCO is regarded as the most abundant enzyme in the biosphere, critical for sustaining life.
5. Match the following enzymes with their substrates:
A. Lipase - I. Urea
B. Urease - II. Lipids
C. Maltase - III. Maltose
D. Amylase - IV. Starch
1. A-II, B-I, C-III, D-IV
2. A-I, B-II, C-III, D-IV
3. A-III, B-IV, C-II, D-I
4. A-IV, B-III, C-I, D-II
Explanation: The correct answer is option 1 (A-II, B-I, C-III, D-IV). Lipase hydrolyzes lipids into fatty acids and glycerol. Urease breaks urea into ammonia and carbon dioxide. Maltase hydrolyzes maltose into glucose, while amylase breaks starch into maltose. These enzymes exemplify substrate specificity and biological efficiency in metabolism.
6. Which of the following is not a characteristic of enzymes?
1. They are specific in their action
2. They are consumed during the reaction
3. They lower activation energy
4. They act as biological catalysts
Explanation: The correct answer is option 2. Enzymes are not consumed during biochemical reactions; they act repeatedly as catalysts. They bind substrates, reduce activation energy, and accelerate reaction rates. After reaction completion, enzymes are regenerated unchanged, allowing them to catalyze multiple cycles without depletion, maintaining metabolic efficiency in living organisms.
7. Fill in the blanks:
_________ catalyzes the conversion of hydrogen peroxide into water and oxygen.
1. Peroxidase
2. Catalase
3. Oxidase
4. Reductase
Explanation: The correct answer is option 2 (Catalase). Catalase decomposes hydrogen peroxide (H₂O₂), a toxic by-product of metabolism, into harmless water and oxygen. Found in peroxisomes of eukaryotic cells, this enzyme protects tissues from oxidative damage and maintains redox homeostasis, making it one of the fastest enzymes known in biological systems.
8. Choose the correct statements:
Statement I: Pepsin works best in acidic pH.
Statement II: Trypsin functions optimally in alkaline pH.
1. Both statements are correct
2. Only Statement I is correct
3. Only Statement II is correct
4. Both statements are incorrect
Explanation: The correct answer is option 1. Pepsin, secreted in the stomach, is active in acidic pH (~2). Trypsin, secreted by the pancreas into the intestine, functions optimally in alkaline pH (~8). These adaptations ensure efficient protein digestion at different regions of the digestive tract, demonstrating enzyme pH specificity and physiological regulation.
9. Which enzyme helps in converting sucrose into glucose and fructose?
1. Invertase
2. Sucrase
3. Lactase
4. Zymase
Explanation: The correct answer is option 1 (Invertase). Invertase catalyzes the hydrolysis of sucrose into glucose and fructose, collectively called invert sugar due to optical rotation change. This enzyme is secreted by yeast and used in confectionery industries for producing soft-centered sweets, demonstrating industrial and biological significance of enzymatic reactions.
10. Which of the following statements about collagen is true?
1. It is an enzyme
2. It is a storage protein
3. It provides tensile strength to connective tissues
4. It is a hormone
Explanation: The correct answer is option 3. Collagen provides tensile strength and structural support to connective tissues such as tendons, ligaments, and skin. It is composed of three polypeptide chains forming a triple helix. This unique structure enables flexibility and durability, making collagen a key component of animal structural integrity and tissue repair.
Topic: Enzymes; Subtopic: Effect of pH on Enzyme Activity
Keyword Definitions:
Enzyme: Biological catalyst that speeds up biochemical reactions by lowering activation energy.
Active site: The specific region on the enzyme where substrate molecules bind and reactions occur.
Optimum pH: The pH at which an enzyme exhibits maximum activity.
Denaturation: Structural alteration of an enzyme leading to loss of function, often caused by extreme pH or temperature.
Substrate: The molecule upon which an enzyme acts to form products.
Lead Question - 2024 (Jhajjhar)
Which of the following graphs explains the effect of pH on the enzyme activity?
Graph 1: Graph 1 shows a bell-shaped curve with enzyme activity on the y-axis and pH on the x-axis.
Graph 2: Graph 2 shows a bell-shaped curve with pH on the y-axis and enzyme activity on the x-axis.
Graph 3: Enzyme activity continuously increases and then plateaus as the pH increases.
Graph 4: Enzyme activity continuously increases and then plateaus as the pH increases.
Explanation:
The correct answer is Graph 1. Enzyme activity typically follows a bell-shaped curve when plotted against pH. Each enzyme has an optimum pH where its activity is maximum. Beyond this range, enzyme structure and active site configuration change due to denaturation, decreasing catalytic efficiency. For example, pepsin works best at acidic pH while trypsin at alkaline pH.
1. The optimum pH for pepsin is approximately:
1. 2
2. 5
3. 7
4. 9
Explanation:
Option 1 is correct. Pepsin, a digestive enzyme found in the stomach, has an optimum pH around 2. It functions best in the highly acidic environment created by hydrochloric acid. Deviations from this pH reduce its catalytic activity as changes in H+ concentration alter the ionization state of amino acids in the active site.
2. Which enzyme functions optimally at alkaline pH?
1. Pepsin
2. Trypsin
3. Amylase
4. Catalase
Explanation:
Option 2 is correct. Trypsin, secreted by the pancreas, works in the small intestine where the environment is slightly alkaline (pH 8). It digests proteins into smaller peptides. The enzyme’s tertiary structure is stable at this pH, while acidic conditions cause denaturation, leading to decreased activity or complete inactivation of the enzyme molecule.
3. When an enzyme is exposed to a very high pH, it:
1. Increases its activity
2. Becomes inactive due to denaturation
3. Remains unaffected
4. Converts into a substrate
Explanation:
Option 2 is correct. Extreme pH conditions disrupt ionic bonds and hydrogen bonds that maintain the enzyme’s tertiary structure. As a result, the enzyme becomes denatured, losing its specific shape and functional active site. Consequently, the substrate can no longer bind efficiently, and the enzyme’s catalytic activity declines or stops entirely.
4. Which of the following statements best explains the effect of pH on enzyme activity?
1. pH alters the enzyme concentration
2. pH affects the shape of the enzyme and active site
3. pH affects temperature
4. pH controls product formation directly
Explanation:
Option 2 is correct. pH modifies the ionization of amino acid residues within the enzyme, especially those forming the active site. This alteration changes the enzyme’s 3D structure, impacting substrate binding. Each enzyme has a specific pH optimum. Deviations either lower or raise H+ concentration, disrupting active site geometry and reducing catalytic efficiency.
5. The enzyme amylase shows maximum activity at:
1. pH 2
2. pH 5
3. pH 6.8
4. pH 9
Explanation:
Option 3 is correct. Salivary amylase, responsible for starch digestion in the mouth, works best around pH 6.8, which is nearly neutral. At very low or high pH values, its activity decreases because the enzyme’s structure is altered. This ensures that enzymatic reactions occur efficiently in specific physiological environments within the human body.
6. Enzyme activity generally decreases beyond optimum pH because:
1. Substrate concentration decreases
2. Active site changes due to altered ionization
3. Enzyme quantity reduces
4. Temperature increases
Explanation:
Option 2 is correct. Changes in pH alter the charge and ionization state of amino acid residues, particularly those in the active site. This leads to conformational changes in the enzyme, reducing substrate affinity and catalytic efficiency. Thus, beyond optimum pH, enzyme activity rapidly falls due to structural distortion or complete denaturation.
Assertion-Reason Type Question
7. Assertion (A): Each enzyme has an optimum pH.
Reason (R): At optimum pH, the enzyme’s active site is most suitable for substrate binding.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, R is false.
4. A is false, R is true.
Explanation:
Option 1 is correct. Every enzyme exhibits maximum activity at a specific pH called the optimum pH. At this value, the active site conformation and substrate binding capability are ideal, ensuring maximum catalytic efficiency. Any deviation changes the enzyme’s charge distribution, lowering substrate affinity and reaction rate, thereby decreasing enzymatic performance significantly.
Matching Type Question
8. Match the enzyme with its optimum pH:
A. Pepsin → i. 7
B. Trypsin → ii. 8
C. Amylase → iii. 6.8
D. Catalase → iv. 2
1. A-iv, B-ii, C-iii, D-i
2. A-i, B-iii, C-iv, D-ii
3. A-ii, B-iv, C-i, D-iii
4. A-iii, B-ii, C-i, D-iv
Explanation:
Option 1 is correct. Pepsin works best at pH 2, trypsin at pH 8, amylase at pH 6.8, and catalase at pH 7. Each enzyme functions optimally at a specific hydrogen ion concentration, reflecting its physiological environment. This precise adaptation ensures maximum efficiency in digestion and metabolism under varying body conditions.
Fill in the Blanks Question
9. The pH value at which an enzyme shows maximum activity is called ______.
1. Activation pH
2. Neutral pH
3. Optimum pH
4. Enzyme pH
Explanation:
Option 3 is correct. The optimum pH is the specific hydrogen ion concentration where an enzyme’s catalytic activity is highest. At this pH, the enzyme’s active site has the most suitable conformation for substrate binding. Extreme deviations in pH disturb active site ionization, altering enzyme shape and reducing catalytic efficiency in biochemical reactions.
Choose the Correct Statements Question
10. Statement I: All enzymes have the same optimum pH.
Statement II: Optimum pH varies depending on enzyme type and its environment.
1. Both statements are true.
2. Both statements are false.
3. Statement I is true, Statement II is false.
4. Statement I is false, Statement II is true.
Explanation:
Option 4 is correct. Optimum pH is not the same for all enzymes; it depends on their source and function. Pepsin acts best at acidic pH, while trypsin prefers alkaline conditions. Variation in enzyme structure, habitat, and function determine the specific pH range for maximum catalytic activity, ensuring physiological balance and efficiency.
Topic: Proteins and Secondary Metabolites; Subtopic: Plant-Derived Compounds and Their Functions
Keyword Definitions:
• Toxin: A poisonous substance produced by living organisms like plants, animals, or microbes that can harm other organisms.
• Polymeric substance: A large molecule made up of repeating monomer units, such as starch, cellulose, or gum.
• Lectin: A protein that binds specifically to carbohydrates and is often involved in cell recognition and defense mechanisms.
• Drug: A natural or synthetic compound used to diagnose, treat, or prevent diseases in living organisms.
• Concanavalin A: A type of lectin obtained from jack beans that binds to glucose and mannose residues.
• Ricin: A highly toxic protein derived from castor beans, inhibiting protein synthesis in cells.
Lead Question – 2024 (Jhajjar)
Match List-I with List-II
List-I | List-II
A. Toxin | I. Gum
B. Polymeric substance | II. Concanavalin A
C. Lectin | III. Ricin
D. Drug | IV. Vinblastin
1. A-III, B-IV, C-II, D-I
2. A-II, B-IV, C-I, D-III
3. A-III, B-I, C-II, D-IV
4. A-II, B-I, C-IV, D-III
Explanation: The correct answer is 3. A-III, B-I, C-II, D-IV. Toxin (Ricin) is a poisonous protein, polymeric substances like gums are plant exudates, Concanavalin A is a lectin that binds carbohydrates, and Vinblastin is a drug used in cancer therapy. Each substance plays a distinct role in biological and medicinal processes.
1. Which of the following is a protein that binds to carbohydrates and is used in cell recognition studies?
1. Ricin
2. Concanavalin A
3. Cellulose
4. Histone
Explanation: Concanavalin A is a lectin that specifically binds to glucose and mannose residues. It is widely used in biochemistry to study glycoproteins and cell surface carbohydrates. Lectins like Concanavalin A play a significant role in immune recognition, plant defense, and cell adhesion processes.
2. Which plant produces the toxin Ricin?
1. Abrus precatorius
2. Ricinus communis
3. Catharanthus roseus
4. Glycine max
Explanation: Ricin is a deadly protein toxin extracted from the seeds of Ricinus communis (castor plant). It inhibits protein synthesis by inactivating ribosomes, causing severe cellular damage. Even small doses can be fatal. Despite its toxicity, the castor plant has industrial uses for oil extraction.
3. Vinblastin, a chemotherapeutic drug, is extracted from which plant?
1. Catharanthus roseus
2. Digitalis purpurea
3. Papaver somniferum
4. Atropa belladonna
Explanation: The anticancer drug Vinblastin is obtained from Catharanthus roseus (Madagascar periwinkle). It inhibits cell division by preventing spindle formation during mitosis. Vinblastin and vincristine are vital alkaloids used in treating cancers like leukemia and Hodgkin’s lymphoma. Their discovery revolutionized plant-based chemotherapy.
4. Which of the following pairs is correctly matched?
1. Morphine – Antibiotic
2. Quinine – Antimalarial
3. Ricin – Hormone
4. Concanavalin A – Lipid
Explanation: The correct pair is Quinine – Antimalarial. Quinine is a natural alkaloid extracted from the bark of the Cinchona tree. It interferes with parasite metabolism and is effective against malaria-causing Plasmodium species. Quinine was one of the first plant-derived drugs used in medical history.
5. Which of the following polymeric substances is secreted by plants for protection and wound healing?
1. Gum
2. Latex
3. Resin
4. Alkaloid
Explanation: Gum is a polymeric substance secreted by plants in response to injury. It consists mainly of polysaccharides and helps seal wounds and prevent microbial invasion. Examples include gum arabic and gum tragacanth. These are also commercially valuable as stabilizers and emulsifiers in industries.
6. Which among the following is an example of an alkaloid?
1. Quinine
2. Gum arabic
3. Pectin
4. Concanavalin A
Explanation: Quinine is an alkaloid derived from the bark of Cinchona trees. Alkaloids are nitrogen-containing organic compounds with pharmacological effects. They often serve as defense molecules in plants and are widely used as drugs, stimulants, and poisons. Examples include morphine, nicotine, and caffeine.
7. Assertion-Reason Type:
Assertion (A): Vinblastin inhibits cell division by interfering with spindle fiber formation.
Reason (R): Vinblastin stabilizes microtubules during mitosis.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, but R is false.
4. Both A and R are false.
Explanation: The correct answer is 3. A is true, but R is false. Vinblastin prevents polymerization of microtubules, thereby blocking spindle formation and arresting mitosis. It does not stabilize microtubules; rather, it depolymerizes them, leading to cell cycle arrest during metaphase. This property makes it effective in cancer treatment.
8. Matching Type:
Match the plant product with its chemical nature:
A. Morphine | I. Alkaloid
B. Gum arabic | II. Polysaccharide
C. Ricin | III. Protein
D. Anthocyanin | IV. Pigment
1. A-I, B-II, C-III, D-IV
2. A-II, B-I, C-IV, D-III
3. A-III, B-I, C-II, D-IV
4. A-IV, B-III, C-I, D-II
Explanation: The correct match is 1. A-I, B-II, C-III, D-IV. Morphine is an alkaloid, gum arabic is a polysaccharide, ricin is a protein, and anthocyanins are pigments responsible for colors in flowers and fruits. Each compound represents a distinct biochemical class found in plant tissues.
9. Fill in the Blanks:
The toxic protein extracted from castor seeds is known as ________.
1. Concanavalin A
2. Ricin
3. Vinblastin
4. Pectin
Explanation: The correct answer is Ricin. It is a ribosome-inactivating protein that blocks protein synthesis in animal cells. Even minute doses can be lethal. Despite its toxicity, Ricinus communis (castor plant) is cultivated for castor oil, which is widely used in industry and medicine.
10. Choose the Correct Statements:
Statement I: Lectins are proteins that bind to specific sugars on cell surfaces.
Statement II: Ricin is an example of a lectin that is non-toxic.
1. Both statements are correct.
2. Only Statement I is correct.
3. Only Statement II is correct.
4. Both statements are incorrect.
Explanation: The correct answer is 2. Only Statement I is correct. Lectins are sugar-binding proteins that play roles in cell signaling and defense. Ricin, however, is a toxic lectin that inhibits protein synthesis. Therefore, while Statement I is true, Statement II is false due to Ricin’s high toxicity.
Topic: Carbohydrate Metabolism; Subtopic: Glycolysis
Keyword Definitions:
• Glycolysis: The breakdown of glucose into pyruvate releasing energy.
• Phosphorylation: Addition of a phosphate group to a molecule.
• Isomerisation: Conversion of one molecule into another with the same formula but different structure.
• ATP: Adenosine triphosphate, the energy currency of the cell.
• Enzyme: A biological catalyst that speeds up biochemical reactions.
• Substrate-level phosphorylation: Direct synthesis of ATP from ADP during metabolism.
Lead Question – 2024 (Jhajjar)
In glycolysis, the conversion of 1,3-bisphosphoglyceric acid to 3-phosphoglyceric acid is:
1. Energy yielding process
2. Energy utilising process
3. Phosphorylation process
4. Isomerisation process
Explanation: The conversion of 1,3-bisphosphoglyceric acid to 3-phosphoglyceric acid is an energy-yielding process catalyzed by the enzyme phosphoglycerate kinase. During this step, ATP is synthesized from ADP through substrate-level phosphorylation. It represents one of the two ATP-generating reactions in glycolysis, producing energy directly within the cytoplasm.
1. Which enzyme catalyzes the first phosphorylation reaction in glycolysis?
1. Hexokinase
2. Phosphofructokinase
3. Aldolase
4. Enolase
Explanation: The first phosphorylation in glycolysis is catalyzed by hexokinase, which converts glucose into glucose-6-phosphate using one ATP molecule. This irreversible reaction traps glucose inside the cell, committing it to the glycolytic pathway. It ensures that glucose remains available for subsequent energy-yielding steps in metabolism.
2. The enzyme enolase catalyzes the conversion of:
1. 2-phosphoglycerate to phosphoenolpyruvate
2. 3-phosphoglycerate to 2-phosphoglycerate
3. Fructose-1,6-bisphosphate to glyceraldehyde-3-phosphate
4. Glucose to glucose-6-phosphate
Explanation: Enolase catalyzes the dehydration of 2-phosphoglycerate to phosphoenolpyruvate (PEP). This step introduces a high-energy enol phosphate bond, preparing the substrate for ATP generation. It’s a crucial preparatory step before pyruvate formation and energy release in the final stages of glycolysis.
3. Which of the following steps in glycolysis involves oxidation?
1. Glyceraldehyde-3-phosphate → 1,3-bisphosphoglycerate
2. Glucose → Glucose-6-phosphate
3. Fructose-6-phosphate → Fructose-1,6-bisphosphate
4. Phosphoenolpyruvate → Pyruvate
Explanation: The oxidation step in glycolysis occurs when glyceraldehyde-3-phosphate is converted to 1,3-bisphosphoglycerate. This reaction, catalyzed by glyceraldehyde-3-phosphate dehydrogenase, involves the reduction of NAD⁺ to NADH. It is the only oxidative step in glycolysis and contributes to the cell’s reducing power.
4. The net gain of ATP molecules per molecule of glucose during glycolysis is:
1. 2
2. 4
3. 6
4. 8
Explanation: The net gain of ATP per molecule of glucose during glycolysis is 2. Four ATP molecules are produced through substrate-level phosphorylation, but two are consumed during the preparatory phase. Therefore, the overall energy yield is 2 ATP and 2 NADH per glucose molecule, under anaerobic conditions.
5. Which of the following is not an intermediate of glycolysis?
1. Pyruvate
2. Citrate
3. Fructose-1,6-bisphosphate
4. Phosphoenolpyruvate
Explanation: Citrate is not an intermediate of glycolysis; it belongs to the citric acid cycle. Glycolysis intermediates include glucose-6-phosphate, 3-phosphoglycerate, and phosphoenolpyruvate. Citrate forms when acetyl-CoA combines with oxaloacetate in the Krebs cycle, representing aerobic energy metabolism.
6. The enzyme catalyzing the last step of glycolysis is:
1. Pyruvate kinase
2. Hexokinase
3. Aldolase
4. Phosphofructokinase
Explanation: Pyruvate kinase catalyzes the final step of glycolysis, converting phosphoenolpyruvate (PEP) to pyruvate. This reaction produces one molecule of ATP via substrate-level phosphorylation. It is an irreversible and highly regulated step, ensuring efficient energy extraction from glucose breakdown.
7. Assertion-Reason Type:
Assertion (A): Glycolysis occurs in the cytoplasm of the cell.
Reason (R): Enzymes of glycolysis are located in mitochondria.
1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is not the correct explanation of A.
3. A is true, but R is false.
4. A is false, but R is true.
Explanation: The correct answer is 3. A is true, but R is false. Glycolysis occurs entirely in the cytoplasm, where all enzymes involved in the pathway are present. Mitochondria are involved in aerobic respiration, not glycolysis. Therefore, the assertion is correct while the reason is false.
8. Matching Type:
Match List I with List II
List I – Enzyme | List II – Reaction
A. Hexokinase | I. Glucose → Glucose-6-phosphate
B. Enolase | II. 2-Phosphoglycerate → PEP
C. Aldolase | III. F-1,6-bisphosphate → DHAP + G3P
D. Pyruvate kinase | IV. PEP → Pyruvate
1. A-I, B-II, C-III, D-IV
2. A-II, B-I, C-IV, D-III
3. A-III, B-IV, C-I, D-II
4. A-IV, B-III, C-II, D-I
Explanation: The correct answer is 1. A-I, B-II, C-III, D-IV. Each enzyme in glycolysis catalyzes a specific substrate conversion. Hexokinase initiates glucose phosphorylation, enolase dehydrates 2-phosphoglycerate, aldolase splits F-1,6-bisphosphate, and pyruvate kinase generates ATP in the final step of glycolysis.
9. Fill in the Blanks:
In glycolysis, the enzyme that converts phosphoenolpyruvate to pyruvate is ________.
1. Pyruvate kinase
2. Enolase
3. Phosphoglycerate kinase
4. Hexokinase
Explanation: The enzyme responsible for converting phosphoenolpyruvate to pyruvate is pyruvate kinase. It catalyzes the substrate-level phosphorylation of ADP to ATP. This step marks the end of glycolysis, producing two molecules of pyruvate per glucose molecule and contributing to cellular energy production.
10. Choose the Correct Statements:
Statement I: ATP is consumed in the initial stages of glycolysis.
Statement II: ATP is generated in later stages of glycolysis.
1. Both statements are correct.
2. Only Statement I is correct.
3. Only Statement II is correct.
4. Both statements are incorrect.
Explanation: The correct answer is 1. Both statements are correct. During glycolysis, ATP is used in the phosphorylation of glucose and fructose-6-phosphate, while ATP is generated during the conversion of 1,3-bisphosphoglycerate and phosphoenolpyruvate to their respective products. The balance ensures a net gain of 2 ATP molecules.
Topic: Polysaccharides; Subtopic: Complex Polysaccharides
Keyword Definitions:
Polysaccharides: Large carbohydrate molecules formed by the condensation of many monosaccharides.
Complex Polysaccharides: Structural or storage polysaccharides with branching or diverse monomer composition.
Inulin: A polymer of fructose units found in tubers, used for carbohydrate storage.
Chitin: A nitrogen-containing polysaccharide forming the exoskeleton of arthropods.
Glucosamine: A monosaccharide derivative forming structural polymers like chitin.
N-acetyl galactosamine: An amino sugar found in glycoproteins and glycolipids.
Lead Question - 2024 (Jhajjhar)
Which of the following is not a complex polysaccharide?
1. Inulin
2. Chitin
3. Glucosamine
4. N-acetyl galactosamine
Explanation: Glucosamine is not a complex polysaccharide; it is a monosaccharide derivative containing an amino group. Inulin and chitin are polymeric carbohydrates with structural and storage functions, while N-acetyl galactosamine is an amino sugar unit in glycoproteins. Hence, Glucosamine is not a complex polysaccharide. (Answer: 3)
1. Which of the following polysaccharides serves as a major structural component in fungi cell walls?
1. Cellulose
2. Chitin
3. Glycogen
4. Starch
Explanation: The fungal cell wall is composed of chitin, a polymer of N-acetylglucosamine, providing rigidity and strength. It resembles cellulose structurally but contains nitrogen, differentiating it from plant polysaccharides. Hence, Chitin is the correct answer. (Answer: 2)
2. Which polysaccharide is stored in the liver and muscles of animals?
1. Glycogen
2. Cellulose
3. Starch
4. Amylose
Explanation: Glycogen is the storage polysaccharide in animals, made of glucose monomers with α-1,4 and α-1,6 linkages. It serves as a readily mobilizable energy reserve in liver and muscle tissues. Hence, the correct answer is Glycogen. (Answer: 1)
3. Inulin is mainly found in which plant organ?
1. Seeds
2. Tubers
3. Leaves
4. Stems
Explanation: Inulin is a fructose polymer found mainly in the tubers of plants like dahlia and chicory. It serves as a carbohydrate reserve and is soluble in water, unlike starch. Hence, Tubers is the correct answer. (Answer: 2)
4. Which of the following polysaccharides is a homopolysaccharide?
1. Starch
2. Chitin
3. Hyaluronic acid
4. Peptidoglycan
Explanation: Homopolysaccharides are composed of only one type of monosaccharide. Starch is made entirely of glucose units and functions as a storage polysaccharide in plants. Chitin, though polymeric, is a nitrogen derivative; hence Starch fits the definition. (Answer: 1)
5. Cellulose and chitin differ in the presence of which group?
1. Hydroxyl group
2. Amino group
3. Carbonyl group
4. Carboxyl group
Explanation: Cellulose is composed of β-D-glucose units, while chitin has N-acetylglucosamine monomers, containing an amino group (-NH₂). This nitrogen substitution gives rigidity to chitin, distinguishing it from cellulose. Hence, Amino group is correct. (Answer: 2)
6. Which of the following is a heteropolysaccharide?
1. Amylose
2. Glycogen
3. Hyaluronic acid
4. Starch
Explanation: Hyaluronic acid is a heteropolysaccharide containing alternating N-acetylglucosamine and glucuronic acid units. It forms part of connective tissues and synovial fluid, giving lubrication and elasticity. Thus, Hyaluronic acid is the correct answer. (Answer: 3)
7. Assertion-Reason Type:
Assertion (A): Starch is a mixture of amylose and amylopectin.
Reason (R): Amylose is a branched polymer while amylopectin is linear.
1. Both A and R are true and R explains A.
2. Both A and R are true but R does not explain A.
3. A is true but R is false.
4. A is false but R is true.
Explanation: Starch indeed contains amylose and amylopectin. However, amylose is unbranched (linear), and amylopectin is branched. Hence, the assertion is true but the reason is false. Correct answer: 3.
8. Matching Type:
Match the following:
A. Glycogen — (i) Animals
B. Cellulose — (ii) Plants
C. Chitin — (iii) Fungi
D. Inulin — (iv) Tubers
1. A-(i), B-(ii), C-(iii), D-(iv)
2. A-(ii), B-(i), C-(iii), D-(iv)
3. A-(i), B-(iii), C-(ii), D-(iv)
4. A-(iv), B-(ii), C-(i), D-(iii)
Explanation: Glycogen stores energy in animals, cellulose in plant walls, chitin in fungi exoskeletons, and inulin in tubers. Thus, the correct match is 1.
9. Fill in the Blanks:
The structural polysaccharide present in arthropod exoskeletons is __________.
1. Cellulose
2. Chitin
3. Pectin
4. Glycogen
Explanation: Chitin, composed of N-acetylglucosamine monomers, provides rigidity and protection to arthropod exoskeletons and fungal cell walls. Hence, the correct answer is Chitin. (Answer: 2)
10. Choose the Correct Statements (Statement I & II):
Statement I: Cellulose is digestible by humans.
Statement II: Cellulose has β-1,4 linkages between glucose units.
1. Both statements are true.
2. Both statements are false.
3. Statement I true, Statement II false.
4. Statement I false, Statement II true.
Explanation: Humans lack cellulase enzyme to hydrolyze β-1,4 linkages in cellulose, making it indigestible. Statement I is false, but Statement II is true. Hence, correct answer is 4.
Topic: Amino Acids and Proteins; Subtopic: Zwitter Ion Nature of Amino Acids
Keyword Definitions:
• Zwitter Ion: A molecule carrying both positive and negative charges but electrically neutral overall.
• Amino Acid: Organic compound containing both amino (-NH₂) and carboxyl (-COOH) groups.
• pH: The hydrogen ion concentration determining acidity or alkalinity of a solution.
• Isoelectric Point: The pH at which an amino acid exists predominantly as a zwitter ion.
• Functional Groups: Reactive parts of molecules that determine their chemical behavior.
Lead Question – 2024 (Jhajjhar)
The following can be found as a zwitter ion:
1. Fatty acid
2. Monosaccharide
3. Amino acid
4. Nucleic acid
Explanation: Amino acids can exist as zwitter ions due to the presence of both acidic (-COOH) and basic (-NH₂) groups. At physiological pH, the carboxyl group donates a proton forming COO⁻, while the amino group accepts a proton forming NH₃⁺. This dual charge keeps the molecule electrically neutral yet internally charged, stabilizing its structure. Answer: Amino acid.
1. Which of the following amino acids is neutral at physiological pH?
1. Lysine
2. Glutamic acid
3. Glycine
4. Arginine
Explanation: Glycine is a neutral amino acid because its side chain is a single hydrogen atom, making it non-polar and uncharged at physiological pH. It still exists as a zwitter ion, carrying both positive and negative charges that balance each other. Answer: Glycine.
2. The isoelectric point of an amino acid is the pH at which:
1. It acts as a strong acid
2. It acts as a strong base
3. It exists as a zwitter ion
4. It ionizes completely
Explanation: The isoelectric point (pI) of an amino acid is the pH at which the molecule carries no net electric charge, existing predominantly as a zwitter ion. At this point, the positive and negative charges balance perfectly, leading to minimal solubility in water. Answer: It exists as a zwitter ion.
3. Which group in an amino acid donates a proton to form the zwitter ion?
1. -NH₂ group
2. -COOH group
3. -OH group
4. -CH₃ group
Explanation: The carboxyl (-COOH) group donates a proton (H⁺) to the amino (-NH₂) group within the same molecule, forming COO⁻ and NH₃⁺. This internal proton transfer results in a zwitter ion, which is stable at physiological pH. Answer: -COOH group.
4. Fill in the Blank:
At physiological pH, an amino acid exists predominantly as a ____________.
1. Cation
2. Anion
3. Zwitter ion
4. Neutral molecule
Explanation: At physiological pH (~7.4), amino acids exist as zwitter ions because their carboxyl group is deprotonated (-COO⁻) and their amino group is protonated (-NH₃⁺). This dual charge keeps the molecule overall neutral yet charged internally. Answer: Zwitter ion.
5. Which of the following statements is correct regarding a zwitter ion?
1. It has equal positive and negative charges.
2. It is an ionic compound.
3. It is insoluble in water.
4. It carries a net positive charge.
Explanation: A zwitter ion has equal positive and negative charges, maintaining overall electrical neutrality. The internal ionic nature enhances solubility in water. Hence, amino acids as zwitter ions remain stable and reactive in aqueous biological environments. Answer: It has equal positive and negative charges.
6. Choose the correct statements:
Statement I: Zwitter ions are amphoteric.
Statement II: They act only as acids.
1. Both statements are true
2. Both statements are false
3. Statement I true, Statement II false
4. Statement I false, Statement II true
Explanation: Zwitter ions are amphoteric because they can act as both acids (donate protons) and bases (accept protons). Thus, they help maintain pH balance in biological systems. Answer: Statement I true, Statement II false.
7. Assertion-Reason:
Assertion (A): Amino acids exist as zwitter ions in solution.
Reason (R): Because they contain both acidic and basic groups in the same molecule.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation
3. A is true but R is false
4. A is false but R is true
Explanation: Both assertion and reason are true, and the reason correctly explains the assertion. The coexistence of acidic (-COOH) and basic (-NH₂) groups enables internal proton transfer, forming a zwitter ion. Answer: Option 1.
8. Matching Type:
Match the amino acid with its characteristic.
A. Glycine → I. Simplest amino acid
B. Alanine → II. Contains methyl side chain
C. Lysine → III. Basic amino acid
D. Glutamic acid → IV. Acidic amino acid
1. A-I, B-II, C-III, D-IV
2. A-II, B-I, C-IV, D-III
3. A-III, B-IV, C-I, D-II
4. A-IV, B-III, C-II, D-I
Explanation: Glycine is the simplest amino acid. Alanine has a methyl side chain. Lysine is basic, while Glutamic acid is acidic due to its carboxyl side chain. Answer: A-I, B-II, C-III, D-IV.
9. Which of the following amino acids has a polar side chain?
1. Serine
2. Valine
3. Leucine
4. Alanine
Explanation: Serine possesses a hydroxyl (-OH) group in its side chain, making it polar and capable of hydrogen bonding. This property enhances solubility and interaction within proteins. Answer: Serine.
10. The zwitter ionic form of an amino acid contains:
1. -NH₂ and -COOH groups
2. -NH₃⁺ and -COOH groups
3. -NH₂ and -COO⁻ groups
4. -NH₃⁺ and -COO⁻ groups
Explanation: In its zwitter ion form, the amino group becomes protonated (-NH₃⁺) and the carboxyl group becomes deprotonated (-COO⁻). These opposite charges balance internally, creating an overall neutral but dipolar molecule. Answer: -NH₃⁺ and -COO⁻ groups.
Topic: Enzymes; Subtopic: Properties and Nature of Enzymes
Keyword Definitions:
Enzymes: Biological catalysts that speed up biochemical reactions in living organisms.
Substrate Specificity: The property of an enzyme to act only on a specific substrate.
Thermophilic Organisms: Microorganisms that thrive at high temperatures, often above 60°C.
Metal Ions: Cofactors such as Zn²⁺, Mg²⁺, or Fe²⁺ that assist enzyme function.
Lead Question – 2024 (Jhajjhar)
Which of the following statement is incorrect about enzymes?
(1) They are highly substrate specific
(2) In thermophilic organisms enzymes can catalyze reactions at high temperatures i.e. 90°C
(3) All enzymes are proteinaceous in nature
(4) Some enzymes have metal ions
Explanation: The incorrect statement is (3). Not all enzymes are purely proteinaceous; some, like ribozymes, are RNA molecules that act as catalysts. Enzymes show substrate specificity, and thermophilic enzymes can function at high temperatures. Metal ions often act as cofactors to enhance enzyme activity. Hence, ribozymes are exceptions to protein-only enzymes.
1. Which enzyme converts glucose to glucose-6-phosphate during glycolysis?
(1) Hexokinase
(2) Phosphofructokinase
(3) Aldolase
(4) Pyruvate kinase
Explanation: The correct answer is (1). Hexokinase catalyzes the phosphorylation of glucose to form glucose-6-phosphate, the first step in glycolysis. This reaction requires ATP and ensures that glucose remains inside the cell. The enzyme’s high affinity for glucose makes it crucial in maintaining blood sugar levels under normal conditions.
2. Which of the following is a non-protein enzyme?
(1) Catalase
(2) Pepsin
(3) Ribozyme
(4) Urease
Explanation: The correct answer is (3). Ribozymes are RNA molecules that catalyze specific biochemical reactions, similar to protein enzymes. They are found in ribosomes where they assist in peptide bond formation during protein synthesis. This discovery proved that not all enzymes are made of proteins, expanding our understanding of catalysis.
3. Which factor does NOT affect enzyme activity?
(1) Temperature
(2) pH
(3) Light intensity
(4) Substrate concentration
Explanation: The correct answer is (3). Enzyme activity is influenced by temperature, pH, and substrate concentration but not by light intensity. Enzymes are biological catalysts that function best at optimal pH and temperature, while deviations can denature the enzyme and reduce its catalytic efficiency.
4. Enzymes increase the rate of a chemical reaction by:
(1) Increasing activation energy
(2) Decreasing activation energy
(3) Increasing the substrate concentration
(4) Changing equilibrium
Explanation: The correct answer is (2). Enzymes work by lowering the activation energy required for a reaction to occur, thus speeding up the rate without being consumed in the process. They do not alter the equilibrium of the reaction but enable it to reach equilibrium faster through efficient catalysis.
5. Coenzymes are generally derived from:
(1) Proteins
(2) Vitamins
(3) Minerals
(4) Lipids
Explanation: The correct answer is (2). Coenzymes are organic molecules derived mainly from vitamins, such as NAD⁺ from niacin and FAD from riboflavin. They assist enzymes in catalyzing reactions by carrying chemical groups or electrons between molecules, thus playing a vital role in metabolism.
6. The enzyme that hydrolyses starch into maltose is:
(1) Lipase
(2) Maltase
(3) Amylase
(4) Sucrase
Explanation: The correct answer is (3). Amylase breaks down starch into maltose units, aiding in carbohydrate digestion. It is present in saliva (salivary amylase) and pancreatic juice (pancreatic amylase). This enzymatic action ensures the conversion of complex polysaccharides into absorbable sugars during digestion.
7. (Assertion-Reason Type)
Assertion (A): Enzymes act as biological catalysts.
Reason (R): Enzymes increase activation energy to speed up reactions.
(1) Both A and R are true, and R is the correct explanation of A
(2) Both A and R are true, but R is not the correct explanation of A
(3) A is true, but R is false
(4) A is false, but R is true
Explanation: The correct answer is (3). Enzymes indeed act as biological catalysts, but they lower—not increase—the activation energy. This reduction allows reactions to proceed more rapidly under physiological conditions, enabling life-sustaining biochemical processes in living organisms.
8. (Matching Type)
Match List I with List II:
List I - Enzyme
A. Lipase
B. Pepsin
C. Amylase
D. Urease
List II - Substrate
I. Protein
II. Fat
III. Starch
IV. Urea
(1) A-II, B-I, C-III, D-IV
(2) A-I, B-II, C-IV, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-III, C-I, D-IV
Explanation: The correct answer is (1). Lipase acts on fats, pepsin on proteins, amylase on starch, and urease on urea. Each enzyme is specific to its substrate due to the complementary structure of the active site, maintaining the principle of enzyme specificity in biological systems.
9. (Fill in the Blank)
The enzyme that converts hydrogen peroxide (H₂O₂) into water and oxygen is ________.
(1) Catalase
(2) Oxidase
(3) Peroxidase
(4) Dehydrogenase
Explanation: The correct answer is (1). Catalase decomposes hydrogen peroxide into water and oxygen, preventing oxidative damage to cells. This enzyme is found in peroxisomes and protects living tissues from toxic effects of accumulated H₂O₂, ensuring cellular metabolic balance.
10. (Choose the Correct Statements)
Statement I: Enzymes are denatured at high temperatures.
Statement II: Enzymes change the equilibrium position of reactions.
(1) Both statements are true
(2) Both statements are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Explanation: The correct answer is (3). Enzymes are sensitive to temperature; high heat denatures them, destroying their catalytic properties. However, enzymes do not change the equilibrium of a reaction—they only accelerate the rate at which equilibrium is achieved by lowering activation energy.
Topic: Enzymes; Subtopic: Mechanism of Enzyme Action
Keyword Definitions:
• Enzyme: Biological catalyst that speeds up biochemical reactions without being consumed.
• Active Site: Region of enzyme where substrate molecules bind and undergo chemical reaction.
• Substrate: Reactant molecule on which an enzyme acts.
• Enzyme-Substrate Complex: Temporary intermediate formed when enzyme binds to its substrate.
• Product: Molecule formed after enzyme catalysis.
Lead Question - 2024
Regarding catalytic cycle of an enzyme action, select the correct sequential steps:
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
(1) A,E,B,D,C
(2) B,A,C,D,E
(3) E,D,C,B,A
(4) E,A,D,C,B
Answer: (4) E,A,D,C,B
Explanation: Enzyme action follows a specific catalytic sequence. The substrate first binds to the active site (E), forming an enzyme-substrate complex (A). The substrate’s chemical bonds are then broken or rearranged (D), resulting in the release of products (C). Finally, the enzyme becomes free (B) to bind with another substrate, repeating the cycle efficiently.
1. Which of the following models best explains enzyme-substrate interaction?
(1) Lock and Key Model
(2) Fluid Mosaic Model
(3) Zwitterion Model
(4) Endosymbiotic Model
Answer: (1) Lock and Key Model
Explanation: The Lock and Key Model explains enzyme specificity, where the substrate fits precisely into the enzyme’s active site like a key into a lock. This structural complementarity ensures that only specific substrates undergo catalysis, providing high precision and efficiency in metabolic processes within living cells.
2. The induced-fit model of enzyme action suggests that:
(1) Enzyme shape remains fixed during catalysis.
(2) Substrate shape changes to fit enzyme.
(3) Enzyme adjusts its shape to fit substrate.
(4) No structural change occurs in enzyme or substrate.
Answer: (3) Enzyme adjusts its shape to fit substrate.
Explanation: The induced-fit model proposed by Daniel Koshland states that enzymes are flexible structures. When a substrate binds, the enzyme undergoes conformational changes to form an optimal fit. This dynamic adjustment lowers activation energy, enhances specificity, and facilitates catalytic reactions vital for metabolism and cellular energy regulation.
3. Which part of enzyme catalysis is directly responsible for lowering activation energy?
(1) Product release
(2) Enzyme-substrate complex formation
(3) Substrate diffusion
(4) Enzyme denaturation
Answer: (2) Enzyme-substrate complex formation
Explanation: The enzyme-substrate complex stabilizes the transition state of the reaction, reducing activation energy. This allows reactants to convert into products faster under mild biological conditions. The enzyme’s specific orientation and microenvironment promote bond rearrangements, ensuring rapid catalysis and maintaining energy efficiency within biological systems.
4. Which of the following factors does not affect enzyme activity?
(1) Temperature
(2) pH
(3) Enzyme concentration
(4) Colour of substrate
Answer: (4) Colour of substrate
Explanation: Enzyme activity is influenced by factors such as temperature, pH, substrate concentration, and enzyme concentration. These parameters affect molecular collisions and enzyme structure. However, the substrate’s colour has no biochemical significance in determining enzyme efficiency or catalytic rate, since enzyme specificity depends on molecular structure, not visual appearance.
5. Enzyme activity doubles with every 10°C rise in temperature until a limit is reached. Beyond that, activity declines due to:
(1) Enzyme activation
(2) Substrate denaturation
(3) Enzyme denaturation
(4) Increased viscosity
Answer: (3) Enzyme denaturation
Explanation: Enzymes are protein molecules sensitive to heat. A moderate rise in temperature increases molecular movement, enhancing enzyme activity. However, excessive heat disrupts hydrogen and ionic bonds in the enzyme’s structure, leading to denaturation and loss of catalytic function. This irreversible process alters the active site, stopping substrate binding.
6. Which of the following enzymes catalyzes the conversion of starch to maltose?
(1) Maltase
(2) Amylase
(3) Lipase
(4) Sucrase
Answer: (2) Amylase
Explanation: Amylase catalyzes the hydrolysis of starch into maltose by breaking α-1,4-glycosidic bonds. It is secreted in saliva and pancreatic juice, initiating carbohydrate digestion. This enzyme functions best around neutral pH and optimum temperature, ensuring effective starch breakdown into simpler sugars for energy production during metabolism.
Assertion-Reason Type Question
7. Assertion (A): Enzymes speed up biochemical reactions.
Reason (R): Enzymes provide energy to start the reaction.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, but R is false.
(4) A is false, but R is true.
Answer: (3) A is true, but R is false.
Explanation: Enzymes accelerate reactions not by providing energy but by lowering the activation energy required. They create a favorable environment for bond breaking and formation, thereby speeding up reaction rates. The reactants’ intrinsic energy remains unchanged; only the pathway to product formation becomes energetically efficient.
Matching Type Question
8. Match the enzymes with their respective substrates:
A. Lipase 1. Proteins
B. Amylase 2. Lipids
C. Pepsin 3. Starch
(1) A-1, B-2, C-3
(2) A-2, B-3, C-1
(3) A-3, B-2, C-1
(4) A-2, B-1, C-3
Answer: (2) A-2, B-3, C-1
Explanation: Lipase acts on lipids to yield fatty acids and glycerol (A-2). Amylase hydrolyzes starch into maltose and glucose (B-3). Pepsin, a gastric enzyme, breaks down proteins into smaller peptides (C-1). These enzyme-substrate relationships ensure proper digestion of macronutrients essential for growth and metabolic activities.
Fill in the Blanks Type Question
9. The non-protein part of an enzyme essential for its activity is called ________.
(1) Apoenzyme
(2) Cofactor
(3) Substrate
(4) Product
Answer: (2) Cofactor
Explanation: Many enzymes require a non-protein component called a cofactor, which may be a metal ion or an organic coenzyme. It helps in stabilizing the enzyme’s structure or directly participates in the catalytic process. Without cofactors, certain enzymes remain inactive, forming an apoenzyme-cofactor complex called a holoenzyme for activity.
Choose the Correct Statements Type Question
10. Statement I: Enzyme activity is influenced by pH and temperature.
Statement II: Denaturation of enzyme increases its catalytic efficiency.
(1) Both statements are true.
(2) Both statements are false.
(3) Statement I is true, Statement II is false.
(4) Statement I is false, Statement II is true.
Answer: (3) Statement I is true, Statement II is false.
Explanation: Enzymes are highly sensitive to pH and temperature, which affect their structure and activity. Optimum conditions allow maximum reaction rates, while deviations cause denaturation and loss of catalytic function. Denaturation alters the enzyme’s active site, preventing substrate binding and reducing, not enhancing, catalytic efficiency.
Topic: Enzymes; Subtopic: Specificity of Enzymes and Types of Bonds Hydrolyzed
Keyword Definitions:
Enzyme: A biological catalyst that speeds up chemical reactions in living organisms without being consumed.
Lipase: An enzyme that hydrolyzes ester bonds in fats into glycerol and fatty acids.
Nuclease: An enzyme that cleaves phosphodiester bonds in nucleic acids.
Protease: An enzyme that breaks peptide bonds in proteins and polypeptides.
Amylase: An enzyme that hydrolyzes glycosidic bonds in starch to release sugars.
Bond types: Ester, peptide, glycosidic, and phosphodiester bonds are key linkages in biomolecules.
Lead Question – 2024
Match List I with List II
A. Lipase – I. Peptide bond
B. Nuclease – II. Ester bond
C. Protease – III. Glycosidic bond
D. Amylase – IV. Phosphodiester bond
Choose the correct answer from the options given below:
(1) A-III, B-II, C-I, D-IV
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-I, C-III, D-II
(4) A-IV, B-II, C-III, D-I
Explanation: Lipase hydrolyzes ester bonds in fats, Nuclease breaks phosphodiester bonds in nucleic acids, Protease cleaves peptide bonds in proteins, and Amylase hydrolyzes glycosidic bonds in polysaccharides. Hence, the correct matching is (2) A-II, B-IV, C-I, D-III. Each enzyme exhibits substrate specificity and catalyzes only one type of biochemical bond, ensuring efficient metabolic regulation.
Guessed Questions:
1. Which of the following enzymes acts on lipids?
(1) Amylase
(2) Lipase
(3) Protease
(4) Nuclease
Explanation: Lipase catalyzes the hydrolysis of ester bonds in triglycerides, releasing fatty acids and glycerol. It is secreted by the pancreas and plays a vital role in fat digestion. The enzyme acts specifically on lipid molecules, maintaining energy balance. Hence, the correct answer is (2) Lipase, which acts on dietary fats and oils efficiently.
2. Which enzyme hydrolyzes nucleic acids into nucleotides?
(1) Protease
(2) Nuclease
(3) Lipase
(4) Amylase
Explanation: Nucleases break down nucleic acids such as DNA and RNA by hydrolyzing phosphodiester bonds between nucleotides. This enzyme is crucial during digestion and DNA repair mechanisms. It ensures recycling of nucleotides for cellular functions. Therefore, the correct answer is (2) Nuclease, which specifically cleaves phosphodiester linkages within or between nucleic acid chains.
3. Proteases are responsible for breaking which type of bonds?
(1) Peptide bonds
(2) Ester bonds
(3) Glycosidic bonds
(4) Phosphodiester bonds
Explanation: Proteases catalyze the hydrolysis of peptide bonds between amino acids in proteins. These enzymes are essential for protein digestion and turnover in cells. Pepsin, trypsin, and chymotrypsin are major proteases in humans. Hence, the correct answer is (1) Peptide bonds, which link amino acids and are broken by proteolytic enzymes during digestion.
4. Which of the following enzymes converts starch into maltose?
(1) Amylase
(2) Lipase
(3) Nuclease
(4) Protease
Explanation: Amylase breaks α-1,4-glycosidic bonds in starch, producing maltose and dextrins. It is secreted by salivary glands and the pancreas. This enzyme initiates carbohydrate digestion in the mouth and continues in the small intestine. Hence, the correct answer is (1) Amylase, a carbohydrate-hydrolyzing enzyme crucial for polysaccharide breakdown.
5. Which of the following is a characteristic feature of enzymes?
(1) They are used up during reaction
(2) They increase activation energy
(3) They are highly specific
(4) They act on all substrates
Explanation: Enzymes are biocatalysts that are highly specific, acting only on particular substrates due to their unique active site configuration. They lower activation energy and are not consumed during reactions. Hence, the correct answer is (3) They are highly specific, ensuring selective and efficient biochemical reactions within cells and organisms.
6. Which enzyme class is responsible for breaking bonds by adding water molecules?
(1) Hydrolases
(2) Oxidoreductases
(3) Ligases
(4) Transferases
Explanation: Hydrolases catalyze hydrolysis reactions, adding water to break covalent bonds in biomolecules. Examples include proteases, lipases, and nucleases. These enzymes function in digestion and metabolic degradation. Therefore, the correct answer is (1) Hydrolases, which mediate cleavage of macromolecules such as proteins, nucleic acids, and lipids by hydrolytic action.
7. Assertion–Reason Type:
Assertion (A): Enzymes are highly specific in their action.
Reason (R): Each enzyme can catalyze multiple unrelated reactions.
(1) Both A and R are true, and R is the correct explanation of A
(2) Both A and R are true, but R is not the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Explanation: Enzymes possess unique active sites that bind to specific substrates only. Hence, they are highly specific, and each catalyzes a particular reaction. The reason is false because enzymes cannot catalyze multiple unrelated reactions. Therefore, the correct answer is (3) A is true but R is false, highlighting enzyme selectivity in metabolism.
. Matching Type:
Match List I with List II
A. Protease – I. Peptide bond
B. Lipase – II. Ester bond
C. Nuclease – III. Phosphodiester bond
D. Amylase – IV. Glycosidic bond
(1) A-I, B-II, C-III, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-IV, C-II, D-I
Explanation: Protease hydrolyzes peptide bonds in proteins, Lipase acts on ester bonds in fats, Nuclease breaks phosphodiester bonds in nucleic acids, and Amylase acts on glycosidic linkages in carbohydrates. Hence, the correct matching is (1) A-I, B-II, C-III, D-IV, representing the functional classification of hydrolytic enzymes in biological systems.
9. Fill in the Blanks:
________ catalyzes the breakdown of proteins into peptides and amino acids.
(1) Lipase
(2) Protease
(3) Nuclease
(4) Amylase
Explanation: Protease enzymes degrade proteins by hydrolyzing peptide bonds, producing smaller peptides and amino acids. They are essential for digestion and cellular protein recycling. Proteolytic enzymes like trypsin and pepsin perform these reactions efficiently. Hence, the correct answer is (2) Protease, which catalyzes the breakdown of complex proteins into absorbable amino acid units.
10. Choose the Correct Statements:
Statement I: Lipase hydrolyzes glycosidic bonds in starch.
Statement II: Nuclease hydrolyzes phosphodiester bonds in nucleic acids.
(1) Both I and II are true
(2) Both I and II are false
(3) I is true but II is false
(4) I is false but II is true
Explanation: Lipase acts on ester bonds in lipids, not glycosidic bonds, while nucleases break phosphodiester bonds linking nucleotides in DNA and RNA. Hence, Statement I is false and Statement II is true. Therefore, the correct answer is (4) I is false but II is true, accurately describing enzymatic bond specificity.
Topic: Enzymes and Proteins; Subtopic: Functional Biomolecules and Their Roles
Keyword Definitions:
GLUT–1: A glucose transporter protein that enables facilitated diffusion of glucose into cells.
Insulin: A peptide hormone secreted by the β-cells of the pancreas that regulates blood glucose levels.
Trypsin: A proteolytic enzyme that hydrolyzes peptide bonds in proteins during digestion.
Collagen: A structural protein that forms the main component of connective tissues like tendons and skin.
Lead Question – 2024
Match List I with List II
List I List II
A. GLUT–1 I. Hormone
B. Insulin II. Enzyme
C. Trypsin III. Intercellular ground substance
D. Collagen IV. Enables glucose transport into cells
Choose the correct answer from the options given below:
(1) A–I, B–II, C–III, D–IV
(2) A–II, B–III, C–IV, D–I
(3) A–III, B–IV, C–I, D–II
(4) A–IV, B–I, C–II, D–III
Explanation: GLUT–1 facilitates glucose entry into cells, Insulin is a peptide hormone that lowers blood sugar levels, Trypsin is a digestive enzyme secreted as trypsinogen, and Collagen is a structural connective tissue protein. Hence, the correct match is A–IV, B–I, C–II, D–III. This classification reflects key biomolecular functions essential for metabolism and structure. Correct answer: (4).
1. Which of the following enzymes helps in digestion of proteins in the small intestine?
(1) Amylase
(2) Trypsin
(3) Lipase
(4) Maltase
Explanation: Trypsin, secreted by the pancreas in its inactive form (trypsinogen), helps in hydrolyzing peptide bonds in proteins, producing smaller peptides and amino acids. It acts in the duodenum under alkaline conditions and complements pepsin’s function in protein digestion. Hence, the correct answer is (2).
2. Which of the following is a structural protein?
(1) Collagen
(2) Amylase
(3) Insulin
(4) Pepsin
Explanation: Collagen is the most abundant structural protein found in connective tissues, providing tensile strength to tendons, ligaments, and skin. It forms triple-helical fibers and is critical in maintaining tissue integrity. Unlike enzymes, collagen does not catalyze reactions. Correct answer: (1).
3. Which transport mechanism involves GLUT–1?
(1) Active transport
(2) Facilitated diffusion
(3) Osmosis
(4) Endocytosis
Explanation: GLUT–1 mediates facilitated diffusion, enabling glucose transport across the plasma membrane without energy expenditure. It ensures steady glucose uptake in tissues like the brain and red blood cells. This process depends on concentration gradients, not ATP. Correct answer: (2).
4. Which organ secretes Insulin?
(1) Liver
(2) Pancreas
(3) Kidney
(4) Adrenal gland
Explanation: Insulin is secreted by β-cells of the Islets of Langerhans in the pancreas. It promotes glucose uptake by cells and glycogen synthesis, thus lowering blood glucose levels. Deficiency leads to diabetes mellitus. Correct answer: (2).
5. Which amino acid is abundant in collagen?
(1) Tryptophan
(2) Glycine
(3) Methionine
(4) Tyrosine
Explanation: Collagen is rich in Glycine, Proline, and Hydroxyproline, which enable tight triple-helix formation. Glycine, being the smallest amino acid, fits efficiently in the structure, providing stability and strength to connective tissues. Correct answer: (2).
6. Which enzyme activates Trypsinogen into Trypsin?
(1) Pepsin
(2) Enterokinase
(3) Amylase
(4) Lipase
Explanation: Trypsinogen is converted to active Trypsin by Enterokinase (Enteropeptidase) in the small intestine. This activation initiates a cascade of digestive enzyme activations aiding protein digestion. Correct answer: (2).
7. Assertion (A): Insulin facilitates glucose uptake in cells.
Reason (R): GLUT–1 is a transporter that helps glucose enter cells.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, but R is false.
(4) A is false, but R is true.
Explanation: Both statements are true but not directly explanatory. Insulin increases glucose uptake by inducing GLUT–4 in muscle and adipose tissues, while GLUT–1 is insulin-independent. Correct answer: (2).
8. Match the following:
A. Pepsin I. Stomach
B. Trypsin II. Small intestine
C. Amylase III. Mouth
D. Lipase IV. Pancreas
(1) A–I, B–II, C–III, D–IV
(2) A–III, B–I, C–II, D–IV
(3) A–I, B–IV, C–II, D–III
(4) A–II, B–I, C–III, D–IV
Explanation: Pepsin acts in the stomach, Trypsin in the small intestine (secreted by the pancreas), Amylase begins starch digestion in the mouth, and Lipase acts in the pancreas and intestines. Correct answer: (1).
9. Fill in the blanks: Collagen is a __________ protein found abundantly in __________ tissues.
(1) Fibrous, connective
(2) Globular, muscular
(3) Hormonal, epithelial
(4) Structural, nervous
Explanation: Collagen is a fibrous structural protein present in connective tissues like tendons, ligaments, and cartilage. It provides elasticity and tensile strength. Correct answer: (1).
10. Statement I: Insulin acts as a hormone for regulating blood sugar.
Statement II: GLUT–1 acts as an enzyme to digest glucose.
(1) Both statements are true.
(2) Statement I is true, Statement II is false.
(3) Both statements are false.
(4) Statement I is false, Statement II is true.
Explanation: Insulin is a hormone that lowers blood sugar, while GLUT–1 is a transporter protein, not an enzyme. It aids glucose transport but does not digest it. Hence, only Statement I is correct. Correct answer: (2).
Topic: Enzyme Kinetics; Subtopic: Enzyme Inhibition
Keyword Definitions:
Enzyme: A protein that catalyzes biochemical reactions.
Substrate: A molecule upon which an enzyme acts.
Inhibitor: A substance that decreases enzyme activity.
Competitive inhibition: Inhibition where inhibitor resembles substrate and binds to enzyme's active site.
Feedback inhibition: End product inhibits an earlier step in the pathway.
Succinic dehydrogenase: An enzyme of the TCA cycle catalyzing conversion of succinate to fumarate.
Malonate: A competitive inhibitor structurally similar to succinate.
Lead Question – 2024
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
(1) Feedback inhibition
(2) Competitive inhibition
(3) Enzyme activation
(4) Cofactor inhibition
Explanation: Malonate inhibits succinic dehydrogenase by competing with succinate for the enzyme's active site. It does not affect other enzymes or steps in the pathway, and inhibition is reversible. This is a textbook example of competitive inhibition because the inhibitor resembles the substrate structurally and prevents substrate binding. Feedback inhibition involves end-product regulation, which is different. Enzyme activation and cofactor inhibition are unrelated mechanisms. Competitive inhibitors increase the apparent Km without changing Vmax, demonstrating classical enzyme kinetics. (Answer: 2)
1. Single Correct Answer:
Which of the following will increase the rate of a competitively inhibited enzyme reaction?
(1) Increase substrate concentration
(2) Increase inhibitor concentration
(3) Decrease enzyme concentration
(4) Decrease pH
Explanation: In competitive inhibition, the inhibitor competes with substrate for the active site. Increasing substrate concentration can outcompete the inhibitor, restoring enzyme activity. Increasing inhibitor or decreasing enzyme reduces activity, and pH changes can denature enzymes but are not specific to overcoming inhibition. Hence, adding more substrate overcomes competitive inhibition. (Answer: 1)
2. Single Correct Answer:
In competitive inhibition, which of the following kinetic changes is observed?
(1) Vmax decreases, Km unchanged
(2) Km increases, Vmax unchanged
(3) Both Vmax and Km decrease
(4) Both Vmax and Km increase
Explanation: Competitive inhibitors increase the apparent Km because more substrate is needed to reach half-maximal velocity, while Vmax remains unchanged as the inhibition can be overcome by high substrate concentration. Non-competitive inhibition reduces Vmax without affecting Km. This classic kinetic signature distinguishes competitive inhibition. (Answer: 2)
3. Single Correct Answer:
Malonate is structurally similar to which substrate in the TCA cycle?
(1) Fumarate
(2) Oxaloacetate
(3) Succinate
(4) Citrate
Explanation: Malonate resembles succinate structurally and competes for succinic dehydrogenase's active site. By occupying the binding site, it prevents succinate conversion to fumarate. This illustrates substrate mimicry in competitive inhibition. It does not resemble fumarate, oxaloacetate, or citrate. (Answer: 3)
4. Single Correct Answer:
Which type of inhibition is reversible and can be overcome by excess substrate?
(1) Non-competitive inhibition
(2) Competitive inhibition
(3) Feedback inhibition
(4) Irreversible inhibition
Explanation: Competitive inhibition is reversible and substrate-dependent. Excess substrate competes with the inhibitor for the active site, restoring enzyme activity. Non-competitive and irreversible inhibition cannot be overcome by substrate increase. Feedback inhibition regulates pathways but is not necessarily reversible by substrate. (Answer: 2)
5. Single Correct Answer:
Which statement about succinic dehydrogenase is correct?
(1) Catalyzes conversion of succinate to fumarate
(2) Requires ATP as cofactor
(3) Located in cytosol
(4) Irreversibly inhibited by malonate
Explanation: Succinic dehydrogenase is a mitochondrial enzyme of the TCA cycle, catalyzing succinate to fumarate conversion. It uses FAD as cofactor, is located in the inner mitochondrial membrane, and is reversibly inhibited by malonate. ATP is not a cofactor, and inhibition is competitive and reversible. (Answer: 1)
6. Single Correct Answer:
Which factor does NOT affect competitive inhibition?
(1) Substrate concentration
(2) Inhibitor concentration
(3) Temperature within optimal range
(4) Enzyme concentration
Explanation: Competitive inhibition is influenced by substrate and inhibitor concentrations. Within optimal temperature range, enzyme activity is stable; temperature changes affect activity but not the competitive mechanism directly. Enzyme concentration affects reaction rate but not inhibition type. Therefore, temperature within optimal limits does not directly affect competitive inhibition. (Answer: 3)
7. Assertion-Reason:
Assertion (A): Malonate is a competitive inhibitor of succinic dehydrogenase.
Reason (R): It binds irreversibly to the active site of the enzyme.
(1) Both A and R are true, R explains A
(2) Both A and R are true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Malonate is indeed a competitive inhibitor, but it binds reversibly to the active site, not irreversibly. Competitive inhibition is reversible and can be overcome by increasing substrate concentration. Therefore, assertion is true, reason is false. (Answer: 3)
8. Matching Type:
Match Enzyme with Inhibitor Type:
A. Succinic dehydrogenase – (i) Competitive
B. Hexokinase – (ii) Feedback inhibition
C. Acetylcholinesterase – (iii) Irreversible inhibition
Options:
(1) A-i, B-ii, C-iii
(2) A-ii, B-i, C-iii
(3) A-i, B-iii, C-ii
(4) A-iii, B-ii, C-i
Explanation: Succinic dehydrogenase is competitively inhibited by malonate. Hexokinase is regulated via feedback inhibition by glucose-6-phosphate. Acetylcholinesterase inhibitors can be irreversible, binding covalently to the enzyme. Matching accurately reflects these classical enzyme inhibition examples. (Answer: 1)
9. Fill in the Blanks:
Malonate binds to the ___ of succinic dehydrogenase, preventing ___ from binding.
(1) Active site, substrate
(2) Allosteric site, cofactor
(3) Regulatory site, product
(4) Substrate site, coenzyme
Explanation: Malonate competes with succinate for the active site of succinic dehydrogenase. By occupying the active site, it prevents substrate binding. This is the hallmark of competitive inhibition, differentiating it from allosteric or feedback inhibition mechanisms. (Answer: 1)
10. Choose Correct Statements:
Statement I: Competitive inhibitors resemble the substrate structurally.
Statement II: Competitive inhibition increases Km but does not change Vmax.
Options:
(1) Both statements are true
(2) Statement I true, Statement II false
(3) Statement I false, Statement II true
(4) Both statements are false
Explanation: Competitive inhibitors are substrate analogs that bind the active site, increasing apparent Km as more substrate is needed to achieve half-maximal velocity. Vmax remains unchanged because high substrate concentrations can overcome the inhibition. Both statements correctly describe competitive inhibition. (Answer: 1)
Topic: Enzyme Cofactors; Subtopic: Metal Ion Cofactors
Keyword Definitions:
Enzyme: Biological catalysts that speed up chemical reactions without being consumed.
Cofactor: Non-protein chemical compound required for the biological activity of some enzymes.
Carboxypeptidase: Proteolytic enzyme that cleaves amino acids from the carboxyl end of proteins or peptides.
Niacin: Water-soluble vitamin B3 involved in NAD/NADP coenzymes.
Flavin: Component of FAD/FMN, acting as a redox cofactor in enzymes.
Haem: Iron-containing porphyrin group, cofactor for hemoproteins and some enzymes.
Zinc: Metal ion cofactor required for the catalytic activity of some enzymes, including carboxypeptidase.
Metal Ion Cofactor: A metal ion essential for the structural stability or catalytic function of enzymes.
Proteolytic Enzyme: Enzymes that hydrolyze peptide bonds in proteins.
Active Site: Specific region of an enzyme where substrate molecules bind and undergo a chemical reaction.
Hydrolysis: Chemical reaction involving the breaking of a bond using water.
Lead Question - 2024:
The cofactor of the enzyme carboxypeptidase is:
(1) Niacin
(2) Flavin
(3) Haem
(4) Zinc
Explanation: Correct answer is (4) Zinc. Carboxypeptidase is a zinc-dependent protease enzyme that hydrolyzes the carboxyl-terminal amino acid from proteins and peptides. Zinc acts as a cofactor by stabilizing the enzyme-substrate complex and activating a water molecule for nucleophilic attack on the peptide bond. Niacin, flavin, and haem are cofactors for other enzymes, not carboxypeptidase. Zinc’s presence in the active site is critical for catalytic function, and deficiency in zinc impairs proteolytic activity. This illustrates the importance of metal ion cofactors in enzymatic reactions and structural stability of enzymes, particularly in hydrolysis reactions.
Guessed MCQs:
1. Single Correct Answer MCQ: Which metal ion is essential for the catalytic activity of carboxypeptidase?
(A) Iron
(B) Magnesium
(C) Zinc
(D) Calcium
Explanation: Correct answer is (C). Zinc is required at the active site of carboxypeptidase to stabilize the transition state and assist in peptide bond hydrolysis. Other metals like iron, magnesium, or calcium are not involved in this enzyme's catalysis.
2. Single Correct Answer MCQ: The function of carboxypeptidase is:
(A) Remove amino acids from N-terminal
(B) Remove amino acids from C-terminal
(C) Oxidize substrates
(D) Phosphorylate proteins
Explanation: Correct answer is (B). Carboxypeptidase cleaves amino acids sequentially from the C-terminal end of proteins or peptides, a process facilitated by the zinc cofactor in the active site for hydrolysis.
3. Single Correct Answer MCQ: Which cofactor type does carboxypeptidase require?
(A) Organic cofactor
(B) Metal ion cofactor
(C) Vitamin B derivative
(D) Flavin group
Explanation: Correct answer is (B). Carboxypeptidase requires a metal ion, zinc, as a cofactor. Organic cofactors and vitamin derivatives are involved in other enzyme systems, not in this proteolytic reaction.
4. Single Correct Answer MCQ: Deficiency of which metal impairs carboxypeptidase activity?
(A) Sodium
(B) Iron
(C) Zinc
(D) Potassium
Explanation: Correct answer is (C). Zinc deficiency reduces enzymatic hydrolysis of peptide bonds by carboxypeptidase. Sodium, iron, or potassium do not directly affect its activity.
5. Single Correct Answer MCQ: Niacin acts as a cofactor for:
(A) Carboxypeptidase
(B) Dehydrogenases
(C) Lysozyme
(D) Amylase
Explanation: Correct answer is (B). Niacin is part of NAD/NADP and functions as a coenzyme for dehydrogenases, whereas carboxypeptidase requires zinc.
6. Single Correct Answer MCQ: Haem is a cofactor in:
(A) Carboxypeptidase
(B) Catalase
(C) Amylase
(D) Lipase
Explanation: Correct answer is (B). Haem is the prosthetic group in catalase and peroxidases, facilitating redox reactions. Carboxypeptidase instead uses zinc as its cofactor.
7. Assertion-Reason MCQ:
Assertion (A): Carboxypeptidase requires zinc for activity.
Reason (R): Zinc stabilizes the substrate and activates water for peptide bond hydrolysis.
(A) Both A and R are true and R explains A
(B) Both A and R are true but R does not explain A
(C) A is true, R is false
(D) A is false, R is true
Explanation: Correct answer is (A). Zinc is essential for the catalytic activity of carboxypeptidase. It helps stabilize the substrate in the active site and activates a water molecule for nucleophilic attack, directly explaining the enzyme’s function.
8. Matching Type MCQ: Match enzymes with their cofactors:
List-I: (A) Carboxypeptidase (B) Dehydrogenase (C) Catalase (D) Amylase
List-II: (I) Zinc (II) Niacin (III) Haem (IV) No cofactor
Options:
1. I II III IV
2. II I III IV
3. I III II IV
4. III II I IV
Explanation: Correct answer is 1. Carboxypeptidase uses zinc (I), dehydrogenase uses niacin (II), catalase uses haem (III), and amylase does not require a cofactor (IV).
9. Fill in the Blanks / Completion MCQ: The cofactor of carboxypeptidase is _______.
(A) Niacin
(B) Zinc
(C) Flavin
(D) Haem
Explanation: Correct answer is (B). Zinc is essential for the catalytic function of carboxypeptidase, facilitating hydrolysis of peptide bonds by stabilizing the substrate and activating water molecules.
10. Choose the correct statements MCQ:
Statement I: Zinc acts as a cofactor in carboxypeptidase.
Statement II: Niacin is required for carboxypeptidase activity.
Options:
(A) Both I and II are correct
(B) Only I is correct
(C) Only II is correct
(D) Both are incorrect
Explanation: Correct answer is (B). Zinc is the true cofactor for carboxypeptidase. Niacin is involved in dehydrogenase reactions and is not required for carboxypeptidase activity.
Topic: Lipids; Subtopic: Phospholipids and Lecithin
Keyword Definitions:
Lecithin: A phospholipid molecule present in cell membranes, aiding structural integrity and signaling.
Phospholipids: Lipid molecules with glycerol backbone, two fatty acids, and a phosphate group, forming membranes.
Glycerides: Simple fats composed of glycerol and fatty acids, mainly triglycerides.
Carbohydrates: Organic compounds of carbon, hydrogen, and oxygen, serving as energy source.
Amino acids: Building blocks of proteins, containing amino and carboxyl groups.
Small Molecular Weight Compounds: Molecules with relatively low mass compared to macromolecules.
Cell Membrane: Lipid bilayer structure enclosing cytoplasm, maintaining homeostasis.
Hydrophilic Head: Phosphate-containing part of phospholipids that interacts with water.
Hydrophobic Tail: Fatty acid chains in phospholipids, repelling water and forming bilayers.
Organic Compounds: Carbon-based molecules essential for life processes.
Structural Lipids: Lipids contributing to cell membrane stability and function.
Lead Question - 2024:
Lecithin, small molecular weight organic compound found in living tissues, is an example of:
(1) Phospholipids
(2) Glycerides
(3) Carbohydrates
(4) Amino acids
Explanation: Correct answer is (1) Phospholipids. Lecithin is a phospholipid consisting of a glycerol backbone, two fatty acids, and a phosphate group linked to choline. It is amphipathic, having hydrophilic head and hydrophobic tail, which enables formation of cell membranes. It is not a glyceride (triglyceride), carbohydrate, or amino acid. Lecithin also participates in lipid transport, membrane signaling, and emulsification in cells. Its small molecular weight compared to large macromolecules like proteins allows it to integrate efficiently in lipid bilayers. Therefore, it is classified as a key structural phospholipid.
Guessed MCQs:
1. Single Correct Answer MCQ: Lecithin is commonly found in:
(A) Plant and animal cell membranes
(B) Blood plasma only
(C) Muscle fibers only
(D) Nucleic acids
Explanation: Correct answer is (A). Lecithin is abundant in cell membranes of both plants and animals. Its amphipathic nature forms lipid bilayers, providing membrane fluidity, integrity, and selective permeability. It is not exclusive to plasma, muscle, or nucleic acids, but rather integral to membrane structure and cellular function.
2. Single Correct Answer MCQ: The hydrophilic part of lecithin is:
(A) Fatty acid tail
(B) Glycerol backbone
(C) Phosphate-choline head
(D) Carbon chain
Explanation: Correct answer is (C). The phosphate-choline head of lecithin interacts with water, making it hydrophilic. The fatty acid tails are hydrophobic. This amphipathic property allows formation of stable lipid bilayers, critical for cell membrane architecture and biological functions like signaling and transport.
3. Single Correct Answer MCQ: Lecithin aids in:
(A) Protein synthesis
(B) Membrane fluidity
(C) DNA replication
(D) Glycolysis
Explanation: Correct answer is (B). Lecithin, as a phospholipid, contributes to membrane fluidity by maintaining a flexible lipid bilayer. It also aids in vesicle formation, emulsification, and signal transduction. It is not directly involved in protein synthesis, DNA replication, or glycolysis.
4. Single Correct Answer MCQ: Lecithin is classified as:
(A) Triglyceride
(B) Phospholipid
(C) Steroid
(D) Carbohydrate
Explanation: Correct answer is (B). Lecithin is a phospholipid with a glycerol backbone, fatty acid chains, and a phosphate group. Triglycerides lack phosphate groups, steroids have fused rings, and carbohydrates are sugar-based, distinguishing lecithin clearly as a phospholipid.
5. Single Correct Answer MCQ: Lecithin helps in transport of:
(A) Proteins
(B) Lipids
(C) Carbohydrates
(D) Nucleotides
Explanation: Correct answer is (B). Lecithin is key in forming lipoproteins for lipid transport in blood and cellular membranes. Its amphipathic structure allows emulsification of fats. Proteins, carbohydrates, and nucleotides require different transport mechanisms.
6. Single Correct Answer MCQ: The major functional role of lecithin in cells is:
(A) Energy storage
(B) Structural support in membranes
(C) Enzymatic activity
(D) Hormone signaling
Explanation: Correct answer is (B). Lecithin forms structural components of membranes, maintaining bilayer integrity and fluidity. While it participates indirectly in signaling, its primary role is structural, not energy storage or enzymatic activity.
7. Assertion-Reason MCQ:
Assertion (A): Lecithin is essential for membrane formation.
Reason (R): Lecithin is amphipathic with hydrophilic head and hydrophobic tail.
(A) Both A and R are true and R explains A
(B) Both A and R are true but R does not explain A
(C) A is true, R is false
(D) A is false, R is true
Explanation: Correct answer is (A). Lecithin’s amphipathic nature allows it to form stable lipid bilayers, providing membrane structure and selective permeability. Hydrophilic heads face aqueous environments while hydrophobic tails form interior, directly explaining its essential role in membranes.
8. Matching Type MCQ: Match components of lecithin:
List-I: (A) Glycerol backbone (B) Fatty acids (C) Phosphate-choline (D) Hydrophilic head
List-II: (I) Structural scaffold (II) Hydrophobic tail (III) Polar group (IV) Interacts with water
Options: (A) (B) (C) (D)
1. I II III IV
2. I II IV III
3. II I III IV
4. IV III II I
Explanation: Correct answer is 1. Glycerol forms scaffold (I), fatty acids form hydrophobic tails (II), phosphate-choline is polar group (III), and hydrophilic head interacts with water (IV). This explains amphipathic character and membrane role.
9. Fill in the Blanks / Completion MCQ: Lecithin is a ______, important in membranes.
(A) Carbohydrate
(B) Phospholipid
(C) Protein
(D) Steroid
Explanation: Correct answer is (B). Lecithin is a phospholipid consisting of glycerol, fatty acids, and phosphate group. Its amphipathic structure forms bilayers, ensuring membrane integrity, selective permeability, and fluidity. It is not a carbohydrate, protein, or steroid.
10. Choose the correct statements MCQ:
Statement I: Lecithin has hydrophobic tails.
Statement II: Lecithin forms lipid bilayers in membranes.
(A) Only Statement I is correct
(B) Only Statement II is correct
(C) Both I and II are correct
(D) Neither I nor II is correct
Explanation: Correct answer is (C). Lecithin has fatty acid hydrophobic tails and hydrophilic phosphate-choline heads. This amphipathic structure allows formation of bilayers, providing membrane integrity and functional compartmentalization within cells.
Topic: Classification of Biomolecules; Subtopic: Terpenoids, Fatty Acids, Nucleic Acids, Polysaccharides
Keyword Definitions:
• Terpenoids: Naturally occurring organic chemicals derived from five-carbon isoprene units, often used in medicine and fragrances.
• Unsaturated fatty acid: Fatty acids that contain one or more double bonds between carbon atoms in the hydrocarbon chain.
• Nucleic acid: Biopolymers (DNA, RNA) responsible for storing and transmitting genetic information.
• Polysaccharide: Carbohydrates composed of long chains of monosaccharide units linked by glycosidic bonds.
• Codeine: An alkaloid used as a pain reliever and cough suppressant.
• Diterpenes: Terpenoids composed of four isoprene units, often bioactive.
• Ricin: A toxic protein derived from castor bean.
• Concanavalin A: A lectin protein that binds specific sugar molecules.
• Lectin: Proteins that recognize and bind carbohydrates specifically.
• Alkaloid: Nitrogen-containing organic compound with significant physiological effects.
• Bioactive: Compounds that affect biological systems.
Lead Question - 2023 (Manipur)
Match List - I with List - II
List-I List-II
(A) Terpenoids (I) Codeine
(B) Unsaturated fatty acid (II) Diterpenes
(C) Nucleic acid (III) Ricin
(D) Polysaccharide (IV) Concanavalin A
Choose the correct answer from the options given below:
Options: (A) (B) (C) (D)
1. II IV III I
2. II I IV III
3. II III I IV
4. II IV I III
Explanation:
Terpenoids are derived from isoprene units, and diterpenes specifically consist of four isoprene units, matching option II for (A). Unsaturated fatty acids do not correspond to proteins or lectins, so (B) matches I, Codeine. Nucleic acids, like DNA and RNA, are associated with toxic proteins in experimental context such as Ricin, matching III. Polysaccharides interact with lectins, so (D) corresponds to IV, Concanavalin A. Therefore, the correct match is (A)-(II), (B)-(I), (C)-(III), (D)-(IV), which corresponds to option 2.
1. Single Correct Answer MCQ:
Which of the following is a diterpene?
1. Terpenoids
2. Codeine
3. Linoleic acid
4. Concanavalin A
Explanation:
Diterpenes are a type of terpenoid composed of four isoprene units. Terpenoids include monoterpenes, sesquiterpenes, and diterpenes. Codeine is an alkaloid, linoleic acid is a fatty acid, and Concanavalin A is a protein lectin. Correct answer is 1.
2. Single Correct Answer MCQ:
Ricin is classified as:
1. Protein
2. Polysaccharide
3. Terpenoid
4. Fatty acid
Explanation:
Ricin is a toxic protein derived from castor bean seeds, known for inhibiting protein synthesis. It is not a polysaccharide, terpenoid, or fatty acid. Correct answer is 1.
3. Single Correct Answer MCQ:
Concanavalin A is an example of:
1. Alkaloid
2. Lectin
3. Terpenoid
4. Fatty acid
Explanation:
Concanavalin A is a lectin protein that binds specific carbohydrate residues, often used in biochemical research. It is not an alkaloid, terpenoid, or fatty acid. Correct answer is 2.
4. Single Correct Answer MCQ:
Unsaturated fatty acids are characterized by:
1. Single bonds only
2. Double bonds between carbon atoms
3. Nitrogen atoms in chain
4. Ring structure
Explanation:
Unsaturated fatty acids contain one or more double bonds between carbon atoms, causing kinks in the chain. Saturated fatty acids have single bonds only. Nitrogen atoms or ring structures are not characteristic. Correct answer is 2.
5. Single Correct Answer MCQ:
Codeine is classified as:
1. Terpenoid
2. Alkaloid
3. Protein
4. Polysaccharide
Explanation:
Codeine is a nitrogen-containing alkaloid extracted from opium poppy, used as analgesic and antitussive. It is not a terpenoid, protein, or polysaccharide. Correct answer is 2.
6. Single Correct Answer MCQ:
Which biomolecule stores genetic information?
1. Polysaccharide
2. Nucleic acid
3. Terpenoid
4. Fatty acid
Explanation:
Nucleic acids, DNA and RNA, store and transmit genetic information in cells. Polysaccharides, terpenoids, and fatty acids do not serve this function. Correct answer is 2.
7. Assertion-Reason MCQ:
Assertion (A): Polysaccharides interact with lectins like Concanavalin A.
Reason (R): Lectins are proteins that specifically bind carbohydrates.
1. Both A and R are true and R explains A
2. Both A and R are true but R does not explain A
3. A is true but R is false
4. Both A and R are false
Explanation:
Polysaccharides contain carbohydrate units that can bind lectins. Lectins are proteins with carbohydrate-binding specificity. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion. Correct answer is 1.
8. Matching Type MCQ:
Match the biomolecule with its correct example:
A. Terpenoid — (i) Diterpenes
B. Unsaturated fatty acid — (ii) Linoleic acid
C. Nucleic acid — (iii) DNA
D. Polysaccharide — (iv) Starch
1. A-(i), B-(ii), C-(iii), D-(iv)
2. A-(ii), B-(i), C-(iv), D-(iii)
3. A-(iii), B-(iv), C-(ii), D-(i)
4. A-(iv), B-(iii), C-(i), D-(ii)
Explanation:
Terpenoids include diterpenes, unsaturated fatty acids include linoleic acid, nucleic acids include DNA, and polysaccharides include starch. Correct matching is A-(i), B-(ii), C-(iii), D-(iv), which is option 1.
9. Fill in the Blanks MCQ:
The toxic protein derived from castor bean is called _______.
1. Codeine
2. Ricin
3. Concanavalin A
4. Terpene
Explanation:
Ricin is a highly toxic protein derived from castor bean seeds. Codeine is an alkaloid, Concanavalin A is a lectin, and terpene is a class of natural compounds. Correct answer is 2.
10. Choose the correct statements MCQ:
Statement I: Terpenoids include diterpenes.
Statement II: Concanavalin A is a lectin that binds polysaccharides.
1. Statement I only
2. Statement II only
3. Both statements are true
4. Both statements are false
Explanation:
Terpenoids encompass diterpenes, and Concanavalin A is a lectin that specifically binds to carbohydrate residues in polysaccharides. Both statements are accurate. Correct answer is 3.
Topic: Proteins and Hormones; Subtopic: Structure and Function of Insulin
Keyword Definitions:
Insulin: A peptide hormone secreted by pancreatic β-cells that regulates glucose metabolism.
Polypeptide chain: A linear sequence of amino acids linked by peptide bonds forming proteins.
C-peptide: A connecting peptide linking A and B chains in proinsulin, removed to form active insulin.
Disulphide bridges: Covalent bonds between sulfur atoms in cysteine residues stabilizing protein structure.
Prohormone: An inactive precursor of a hormone requiring enzymatic cleavage for activation.
Lead Question - 2023 (Manipur)
Select incorrect statement regarding chemical structure of insulin:
1. Mature insulin molecule consists of three polypeptide chains - A, B and C.
2. Insulin is synthesized as prohormone which contains extra stretch of C-peptide.
3. C-peptide is not present in mature insulin molecule.
4. Polypeptide chains A and B are linked by disulphide bridges.
Explanation: The mature insulin molecule consists of two polypeptide chains (A and B) linked by disulphide bridges. It is synthesized as a prohormone, proinsulin, which includes a C-peptide that connects A and B chains. During maturation, C-peptide is cleaved and not present in active insulin. Hence, statement 1 is incorrect. (Answer: 1)
Guessed Questions:
1. Which organ secretes insulin?
1. Liver
2. Pancreas
3. Adrenal gland
4. Thyroid gland
Explanation: Insulin is secreted by the β-cells of the islets of Langerhans in the pancreas. It regulates blood glucose by promoting uptake in cells and converting excess glucose to glycogen. Its deficiency causes hyperglycemia and diabetes mellitus. Hence, the correct answer is pancreas. (Answer: 2)
2. The number of disulphide bridges in human insulin is:
1. One
2. Two
3. Three
4. Four
Explanation: Human insulin consists of two inter-chain disulphide bonds connecting A and B chains and one intra-chain disulphide bond within the A-chain, totaling three disulphide bridges. These bonds provide stability and biological functionality to the hormone. Thus, the correct answer is three. (Answer: 3)
3. The C-peptide of proinsulin helps in:
1. Formation of active insulin molecule
2. Regulating glucose uptake
3. Forming glycogen
4. Stimulating glucagon release
Explanation: C-peptide connects A and B chains of proinsulin, ensuring proper folding and structural alignment for active insulin formation. It is cleaved during insulin maturation and serves as a clinical marker for endogenous insulin secretion levels. The correct answer is formation of active insulin molecule. (Answer: 1)
4. Assertion (A): Insulin is a peptide hormone.
Reason (R): It is made up of amino acids linked by peptide bonds.
1. Both A and R are true, and R explains A.
2. Both A and R are true, but R does not explain A.
3. A is true, R is false.
4. A is false, R is true.
Explanation: Insulin, being composed of two amino acid chains joined by peptide and disulphide bonds, is indeed a peptide hormone. The reason correctly explains the assertion since peptide hormones are amino acid-based. Therefore, both statements are true, and R correctly explains A. (Answer: 1)
5. Match the following:
A. Insulin – i. Increases blood glucose
B. Glucagon – ii. Decreases blood glucose
C. Epinephrine – iii. Mobilizes glucose
1. A–ii, B–i, C–iii
2. A–i, B–ii, C–iii
3. A–iii, B–ii, C–i
4. A–ii, B–iii, C–i
Explanation: Insulin decreases blood glucose by facilitating glucose uptake. Glucagon increases blood glucose via glycogenolysis, and epinephrine mobilizes glucose during stress. Correct matching: A–ii, B–i, C–iii. Together, they regulate glucose homeostasis in humans. Thus, the correct answer is option 1. (Answer: 1)
6. Insulin mainly acts on:
1. Brain
2. Muscles and liver
3. Lungs
4. Heart
Explanation: Insulin primarily targets liver, muscle, and adipose tissues to promote glucose uptake and storage as glycogen. It also inhibits gluconeogenesis and lipolysis. The brain is insulin-independent for glucose transport. Hence, insulin mainly acts on muscles and liver. (Answer: 2)
7. Fill in the blanks:
Insulin is synthesized in the _______ of the pancreas.
1. Alpha cells
2. Beta cells
3. Delta cells
4. Acinar cells
Explanation: Insulin is synthesized and secreted by β-cells located in the islets of Langerhans of the pancreas. These cells play a key role in glucose homeostasis by releasing insulin in response to elevated blood sugar levels. Hence, the correct answer is beta cells. (Answer: 2)
8. Choose the correct statements:
Statement I: Insulin promotes glycogen breakdown.
Statement II: Insulin facilitates glucose uptake into cells.
1. Both statements are correct.
2. Both statements are incorrect.
3. Statement I is incorrect, Statement II is correct.
4. Statement I is correct, Statement II is incorrect.
Explanation: Insulin facilitates glucose entry into cells, promoting glycogen synthesis, not its breakdown. Glycogen breakdown is stimulated by glucagon. Therefore, Statement I is incorrect while Statement II is correct. Hence, the correct answer is option 3. (Answer: 3)
9. The gene for human insulin was transferred and expressed first in:
1. Yeast
2. E. coli
3. Bacillus subtilis
4. Aspergillus
Explanation: The first recombinant DNA technology-based human insulin, known as Humulin, was produced using Escherichia coli. The human insulin gene was inserted into bacterial plasmid, allowing bacterial cells to synthesize insulin identical to natural form. Hence, the correct answer is E. coli. (Answer: 2)
10. Insulin deficiency leads to which condition?
1. Hypoglycemia
2. Diabetes mellitus
3. Goitre
4. Obesity
Explanation: Insulin deficiency results in impaired glucose utilization, causing elevated blood glucose levels or hyperglycemia, leading to diabetes mellitus. The disorder is characterized by polyuria, polydipsia, and polyphagia. Hence, the correct answer is diabetes mellitus. (Answer: 2)
Subtopic: Secondary Metabolites
Keyword Definitions:
Secondary Metabolites: Organic compounds produced by plants not directly involved in growth, development, or reproduction but often involved in defense, pigmentation, or signaling.
Curcumin: A phenolic compound found in turmeric with medicinal properties; a secondary metabolite.
Morphine: An alkaloid obtained from opium poppy; used as a painkiller; a secondary metabolite.
Anthocyanin: Water-soluble pigment responsible for red, purple, or blue colors in plants; a secondary metabolite.
Lecithin: A phospholipid forming part of cell membranes; primary metabolite involved in structural function.
Primary Metabolites: Compounds essential for growth and development, including sugars, amino acids, nucleotides, and lipids.
Alkaloids: Nitrogen-containing secondary metabolites with pharmacological activity.
Phenolics: Aromatic secondary metabolites involved in defense and pigmentation.
Lead Question – 2023 (Manipur)
Which of the following is not a secondary metabolite?
1. Curcumin
2. Morphine
3. Anthocyanin
4. Lecithin
Explanation: Curcumin, Morphine, and Anthocyanins are all secondary metabolites involved in defense, pigmentation, or medicinal properties. Lecithin, however, is a primary metabolite, a phospholipid essential for cell membrane structure and function. Unlike secondary metabolites, lecithin is directly required for growth and metabolism. Therefore, the compound that is not a secondary metabolite among the given options is Lecithin. Correct answer: 4. Lecithin.
1. Single Correct Answer MCQ:
Which of the following is an alkaloid?
1. Curcumin
2. Morphine
3. Anthocyanin
4. Lecithin
Explanation: Morphine is an alkaloid derived from opium poppy and acts as a secondary metabolite with pharmacological properties. Alkaloids are nitrogen-containing secondary metabolites. Correct answer: 2. Morphine.
2. Single Correct Answer MCQ:
Anthocyanins are classified as:
1. Alkaloids
2. Terpenoids
3. Phenolics
4. Lipids
Explanation: Anthocyanins are water-soluble pigments responsible for red, purple, or blue colors in plants and belong to the phenolic class of secondary metabolites. Correct answer: 3. Phenolics.
3. Single Correct Answer MCQ:
Which of the following is a primary metabolite?
1. Morphine
2. Lecithin
3. Curcumin
4. Anthocyanin
Explanation: Lecithin is a primary metabolite; it is a structural phospholipid essential for cell membrane integrity and growth, unlike secondary metabolites which are not essential for primary growth. Correct answer: 2. Lecithin.
4. Single Correct Answer MCQ:
Curcumin is derived from which plant?
1. Opium poppy
2. Turmeric
3. Beetroot
4. Tea
Explanation: Curcumin is a phenolic secondary metabolite extracted from turmeric (Curcuma longa) and is responsible for its yellow color and medicinal properties. Correct answer: 2. Turmeric.
5. Single Correct Answer MCQ:
Which compound acts as a natural pigment in plants?
1. Morphine
2. Lecithin
3. Anthocyanin
4. Curcumin
Explanation: Anthocyanins are pigments giving red, purple, or blue coloration to flowers and fruits, serving as a secondary metabolite with protective functions. Correct answer: 3. Anthocyanin.
6. Single Correct Answer MCQ:
Which metabolite has medicinal properties as a painkiller?
1. Lecithin
2. Curcumin
3. Morphine
4. Anthocyanin
Explanation: Morphine, an alkaloid secondary metabolite, is derived from opium poppy and is used as a potent analgesic in medicine. Correct answer: 3. Morphine.
7. Assertion-Reason MCQ:
Assertion (A): Lecithin is essential for plant cell structure.
Reason (R): Lecithin is a secondary metabolite involved in pigmentation.
1. Both A and R true, R explains A
2. Both A and R true, R does not explain A
3. A true, R false
4. A false, R true
Explanation: Lecithin is essential for cell membranes (primary metabolite). It is not a secondary metabolite and does not contribute to pigmentation. Correct answer: 3. A true, R false.
8. Matching Type MCQ:
Match metabolite with type:
A. Curcumin – (i) Phenolic
B. Morphine – (ii) Alkaloid
C. Anthocyanin – (iii) Pigment
D. Lecithin – (iv) Primary metabolite
1. A–i, B–ii, C–iii, D–iv
2. A–ii, B–i, C–iv, D–iii
3. A–iii, B–iv, C–i, D–ii
4. A–iv, B–iii, C–ii, D–i
Explanation: Curcumin is phenolic, Morphine is alkaloid, Anthocyanin is pigment, Lecithin is a primary metabolite. Correct option: 1.
9. Fill in the Blanks MCQ:
__________ is a secondary metabolite with pharmacological properties, while __________ is a primary metabolite involved in cell membrane structure.
1. Morphine, Lecithin
2. Lecithin, Morphine
3. Curcumin, Anthocyanin
4. Anthocyanin, Curcumin
Explanation: Morphine is a secondary metabolite with analgesic properties. Lecithin is a primary metabolite, essential for membrane structure. Correct answer: 1.
10. Choose the Correct Statements MCQ:
Statement I: Curcumin is a secondary metabolite.
Statement II: Lecithin is a secondary metabolite.
1. Both true
2. Both false
3. I true, II false
4. I false, II true
Explanation: Curcumin is a secondary metabolite involved in defense and medicinal functions. Lecithin is a primary metabolite essential for membranes. Correct answer: 3.
Chapter: Plant Secondary Metabolites; Topic: Chemical Defense Mechanisms in Plants; Subtopic: Cardiac Glycosides and Herbivore Deterrence
Calotropis: A common weed that secretes a milky latex containing toxic compounds used as chemical defense against herbivores.
Cardiac glycosides: Toxic secondary metabolites affecting heart muscle function, deterring herbivory.
Ricin: A potent toxin derived from castor seeds interfering with protein synthesis.
Quinine: Bitter alkaloid from Cinchona bark used as antimalarial but acts as herbivore deterrent.
Strychnine: Alkaloid toxin from Strychnos nux-vomica seeds acting on nervous system.
Secondary metabolites: Compounds produced by plants not for growth but for defense.
Herbivory: Feeding of animals on plants causing potential damage to vegetation.
Latex: Milky fluid secreted by plants like Calotropis containing defensive chemicals.
Defense mechanism: Strategies evolved by plants to resist or deter predators and herbivores.
Allelochemicals: Biochemicals influencing interactions between organisms, often for defense.
Toxin: Harmful chemical produced naturally by living organisms for protection or predation.
Lead Question - 2023 (Manipur)
For chemical defense against herbivores, Calotropis has:
1. Cardiac glycosides
2. Strychnine
3. Toxic ricin
4. Distasteful quinine
Explanation: Calotropis secretes a milky latex containing cardiac glycosides, toxic chemicals that interfere with the heart’s normal rhythm when ingested by herbivores. This chemical defense discourages grazing and protects the plant. These glycosides act as potent deterrents against most animals. Strychnine, ricin, and quinine are toxic alkaloids found in other plant species but not in Calotropis. Correct answer: 1.
1. Which of the following is an example of a plant chemical defense?
a) Production of nectar
b) Production of cardiac glycosides
c) Rapid leaf fall
d) Sunlight reflection by leaves
Explanation: Many plants, including Calotropis, produce cardiac glycosides as chemical defenses to deter herbivory. These compounds cause nausea or cardiac arrest when consumed by animals. Nectar and leaf fall are not defense mechanisms, and sunlight reflection aids temperature regulation. Such chemicals are key survival adaptations. Correct answer: b.
2. Ricin, a toxic protein interfering with protein synthesis, is produced by:
a) Calotropis
b) Castor plant
c) Cinchona
d) Datura
Explanation: Ricin is a highly toxic protein obtained from castor seeds (Ricinus communis). It inactivates ribosomes, blocking protein synthesis, and is a potent biological toxin. Calotropis produces cardiac glycosides, not ricin, while Cinchona produces quinine and Datura produces tropane alkaloids. Correct answer: b.
3. Which plant produces the alkaloid quinine?
a) Cinchona
b) Calotropis
c) Castor
d) Strychnos
Explanation: Quinine, a bitter alkaloid used as an antimalarial, is derived from the bark of Cinchona trees. It also acts as a deterrent to herbivores due to its bitterness. Calotropis contains cardiac glycosides, castor has ricin, and Strychnos contains strychnine. Correct answer: a.
4. What type of chemical defense does Datura exhibit?
a) Production of glycosides
b) Production of tropane alkaloids
c) Production of flavonoids
d) Production of saponins
Explanation: Datura species produce tropane alkaloids like atropine, hyoscyamine, and scopolamine, which are neurotoxic compounds acting as chemical defenses against herbivores. These alkaloids deter grazing due to toxicity and bitter taste. Glycosides, flavonoids, and saponins occur in other plants but serve different functions. Correct answer: b.
5. Which of the following statements about cardiac glycosides is true?
a) They are proteins that denature easily
b) They are secondary metabolites affecting the nervous system
c) They are steroidal compounds affecting heart function
d) They are carbohydrates used for energy storage
Explanation: Cardiac glycosides are steroidal compounds that inhibit sodium-potassium ATPase in heart tissue, causing cardiac toxicity in herbivores. They act as defensive secondary metabolites, not energy storage molecules. Their bitterness and toxicity prevent animals from feeding on such plants. Correct answer: c.
6. Calotropis latex acts as a chemical defense because it:
a) Attracts herbivores
b) Produces nectar for pollinators
c) Contains toxic cardiac glycosides
d) Helps in water absorption
Explanation: Calotropis latex is rich in cardiac glycosides and proteolytic enzymes that make it highly toxic and bitter to herbivores. These substances cause nausea and heart irregularities in grazing animals, discouraging feeding. The latex plays no role in water uptake or attraction. Correct answer: c.
7. Assertion (A): Calotropis plants are rarely grazed by animals.
Reason (R): They produce toxic cardiac glycosides that act as deterrents.
a) Both A and R are true, and R explains A
b) Both A and R are true, but R does not explain A
c) A is true, R is false
d) A is false, R is true
Explanation: Animals avoid grazing on Calotropis due to the presence of toxic cardiac glycosides in its latex, which can cause vomiting and heart irregularities. Hence, both assertion and reason are true, and R correctly explains A. This adaptation prevents loss of plant biomass to herbivory. Correct answer: a.
8. Match the plant with its major defensive chemical:
List-I: (A) Calotropis, (B) Castor, (C) Datura, (D) Cinchona
List-II: (I) Ricin, (II) Quinine, (III) Tropane alkaloids, (IV) Cardiac glycosides
a) A-IV, B-I, C-III, D-II
b) A-III, B-II, C-IV, D-I
c) A-II, B-IV, C-I, D-III
d) A-I, B-II, C-III, D-IV
Explanation: Calotropis produces cardiac glycosides, Castor produces ricin, Datura synthesizes tropane alkaloids, and Cinchona contains quinine. These represent diverse plant defense mechanisms against herbivores and pathogens. Correct answer: a.
9. Fill in the blank: The toxic substance responsible for Calotropis defense is __________.
a) Quinine
b) Cardiac glycosides
c) Ricin
d) Caffeine
Explanation: Calotropis uses cardiac glycosides for its chemical defense strategy. These glycosides affect the heart muscles and taste bitter, deterring herbivory. Such toxins are stored in milky latex, preventing loss from grazing animals. Quinine, ricin, and caffeine occur in other plants. Correct answer: b.
10. Choose the correct statements:
Statement I: Calotropis produces cardiac glycosides in latex.
Statement II: These compounds make the plant palatable to herbivores.
a) Both I and II are correct
b) I is correct, II is incorrect
c) I is incorrect, II is correct
d) Both I and II are incorrect
Explanation: Calotropis contains cardiac glycosides that render it toxic and unpalatable to herbivores. These compounds discourage grazing and act as an effective plant defense. Therefore, Statement I is correct and Statement II is incorrect. This adaptation allows survival in environments with heavy herbivore presence. Correct answer: b.
Topic: Carbohydrates; Subtopic: Polysaccharides
Inulin: A storage polysaccharide composed mainly of fructose units, found in plants like chicory and garlic.
Fructose: A monosaccharide sugar commonly found in fruits and honey.
Polysaccharide: A carbohydrate polymer made up of many monosaccharide units.
Glucose: A six-carbon monosaccharide sugar used as energy source in cells.
Galactose: A monosaccharide sugar found in milk and dairy products.
Amino Acids: Building blocks of proteins, not carbohydrates.
Chicory: A plant rich in inulin used as a dietary fiber.
Storage Polysaccharide: A carbohydrate stored in plants or animals for energy supply.
Oligosaccharides: Carbohydrates made of 2–10 monosaccharide units.
Prebiotic: Compounds like inulin that promote growth of beneficial gut bacteria.
Dietary Fiber: Plant-derived carbohydrate that resists digestion in the small intestine.
Lead Question - 2023 (Manipur)
Inulin is a polymer of:
1. Fructose
2. Galactose
3. Amino Acids
4. Glucose
Explanation: Inulin is a storage polysaccharide composed mainly of fructose units linked by β-(2→1) glycosidic bonds. It is found in plants like chicory, garlic, and onion and acts as a prebiotic dietary fiber. It is not composed of glucose, galactose, or amino acids. Inulin helps regulate blood sugar and promotes gut health by supporting beneficial bacteria. Correct answer: 1.
1. Which of the following is a storage polysaccharide in plants?
a) Inulin
b) Cellulose
c) Chitin
d) Lactose
Explanation: Inulin is a storage polysaccharide mainly composed of fructose units in plants like chicory and garlic. Cellulose is structural, chitin is structural in fungi and arthropods, and lactose is a disaccharide in milk. Inulin functions as energy storage and prebiotic fiber. Correct answer: a.
2. Fructans are polymers of:
a) Glucose
b) Fructose
c) Galactose
d) Amino acids
Explanation: Fructans, including inulin and levan, are polysaccharides made of fructose units linked by β-(2→1) or β-(2→6) glycosidic bonds. Glucose, galactose, or amino acids are not the building units of fructans. Fructans are stored in roots and tubers and act as prebiotics. Correct answer: b.
3. Inulin is commonly found in:
a) Wheat and rice
b) Garlic and chicory
c) Milk and cheese
d) Egg white
Explanation: Inulin is abundant in garlic, chicory, onion, and Jerusalem artichoke. It is absent in cereals like wheat and rice, dairy products, or animal sources. Its fructose polymer structure allows it to act as prebiotic dietary fiber, supporting beneficial gut microbiota. Correct answer: b.
4. Which property is true for inulin?
a) Highly digestible in small intestine
b) Acts as dietary fiber
c) Composed of amino acids
d) Found in animal tissues
Explanation: Inulin resists digestion in the small intestine and reaches the colon where it acts as dietary fiber and prebiotic. It is composed of fructose, not amino acids, and is present in plants, not animal tissues. Its fiber property helps regulate gut microbiota and glycemic response. Correct answer: b.
5. The type of glycosidic linkage in inulin is:
a) α-(1→4)
b) β-(1→4)
c) β-(2→1)
d) α-(1→6)
Explanation: Inulin has β-(2→1) glycosidic linkages between fructose units, forming a linear polymer. α-(1→4) linkages are in starch, β-(1→4) in cellulose, α-(1→6) in branched polysaccharides. This structure confers resistance to digestion and prebiotic function. Correct answer: c.
6. Which of the following is a prebiotic?
a) Inulin
b) Lactose
c) Glucose
d) Fructose
Explanation: Inulin is a prebiotic because it stimulates growth of beneficial gut bacteria like Bifidobacterium. Lactose, glucose, and fructose are simple sugars digested in the small intestine and do not have prebiotic properties. Inulin supports colon health and regulates microbiota composition. Correct answer: a.
7. Assertion (A): Inulin is a polymer of fructose.
Reason (R): Inulin is digestible in the human small intestine.
a) Both A and R are true, R explains A
b) Both A and R are true, R does not explain A
c) A is true, R is false
d) A is false, R is true
Explanation: Inulin is a fructose polymer (A is true) but resists digestion in the small intestine, acting as dietary fiber and prebiotic. Therefore, Reason is false. Correct answer: c.
8. Match the plant with the polysaccharide it stores:
1. Chicory
2. Potato
3. Milk
4. Onion
a) 1-Inulin, 2-Starch, 3-Lactose, 4-Inulin
b) 1-Starch, 2-Lactose, 3-Inulin, 4-Starch
c) 1-Lactose, 2-Starch, 3-Inulin, 4-Glucose
d) 1-Glucose, 2-Inulin, 3-Starch, 4-Lactose
Explanation: Chicory and onion store inulin, potato stores starch, milk stores lactose. This reflects plant and animal carbohydrate storage patterns. Correct answer: a.
9. Fill in the blank: ____________ is a fructose polymer found in plants like chicory and garlic.
a) Inulin
b) Starch
c) Cellulose
d) Glycogen
Explanation: Inulin is a polymer of fructose found in chicory, garlic, and onion. Starch and glycogen are glucose polymers; cellulose is a structural β-(1→4) glucose polymer. Inulin serves as storage carbohydrate and prebiotic. Correct answer: a.
10. Choose the correct statements:
Statement I: Inulin is a fructose polymer.
Statement II: Inulin is digestible in the human small intestine.
a) Both I and II are correct
b) Only I is correct
c) Only II is correct
d) Both I and II are incorrect
Explanation: Inulin is indeed a fructose polymer (Statement I correct). It resists digestion in the small intestine and acts as dietary fiber (Statement II incorrect). Therefore, only Statement I is correct. Correct answer: b.
Chapter: Biomolecules; Topic: Structure and Function of Biomolecules; Subtopic: Proteins, Carbohydrates, Lipids, and Nucleic Acids
Keyword Definitions:
• Protein: Large biomolecules composed of amino acids linked by peptide bonds, performing structural, enzymatic, and regulatory functions in cells.
• Unsaturated fatty acid: Fatty acids containing one or more C=C double bonds in the hydrocarbon chain, liquid at room temperature.
• Nucleic acid: Polymers of nucleotides containing a phosphate group, pentose sugar, and nitrogenous base; DNA and RNA are examples.
• Polysaccharide: Carbohydrate polymers of monosaccharides linked by glycosidic bonds; function as energy storage or structural components.
• Peptide bond: Covalent bond linking amino acids in proteins.
• Glycosidic bond: Covalent bond connecting monosaccharides in carbohydrates.
• Phosphodiester bond: Covalent bond linking nucleotides in nucleic acids.
• C=C double bond: Unsaturated bond present in fatty acids.
Lead Question - 2023 (Manipur)
Match List - I with List - II
List-I List-II
(A) Protein (I) C=C double bonds
(B) Unsaturated fatty acid (II) Phosphodiester bond
(C) Nucleic acid (III) Glycosidic bonds
(D) Polysaccharide (IV) Peptide bonds
Choose the correct answer from the options given below:
Options: (A) (B) (C) (D)
1. II I IV III
2. IV III I II
3. IV I II III
4. I IV III II
Answer & Explanation: Option 2. Proteins are polymers of amino acids linked by peptide bonds (IV). Unsaturated fatty acids contain C=C double bonds (I). Nucleic acids have nucleotides joined via phosphodiester bonds (II). Polysaccharides consist of monosaccharides linked by glycosidic bonds (III). Understanding the types of chemical bonds in biomolecules is essential for explaining their structure-function relationship. Proteins serve as enzymes and structural elements, fatty acids provide energy storage and membrane fluidity, nucleic acids store genetic information, and polysaccharides act as energy reserves or structural components.
1. Single Correct Answer MCQ
Which of the following amino acids contains a sulfur atom in its side chain?
Options:
A. Lysine
B. Methionine
C. Valine
D. Leucine
Answer & Explanation: Option B. Methionine is a sulfur-containing amino acid important for initiating protein synthesis and acting as a methyl group donor in various biochemical reactions. Lysine, valine, and leucine do not have sulfur. Methionine’s sulfur atom is crucial for disulfide bond formation and maintaining protein tertiary structure, contributing to protein stability and function in enzymatic reactions and metabolic processes. Proper understanding of sulfur amino acids is important for studying protein chemistry and cellular metabolism.
2. Single Correct Answer MCQ
Which polysaccharide serves primarily as a storage form of glucose in animals?
Options:
A. Starch
B. Cellulose
C. Glycogen
D. Chitin
Answer & Explanation: Option C. Glycogen is the main glucose storage polysaccharide in animals, highly branched, allowing rapid mobilization of glucose. Starch is the plant storage form. Cellulose provides structural support in plant cell walls, and chitin is found in arthropod exoskeletons. Glycogen is synthesized in the liver and muscles, providing energy during fasting or exercise. Its branched structure facilitates quick release of glucose for metabolic needs. Understanding glycogen is critical in physiology and biochemistry to explain energy storage and mobilization mechanisms.
3. Single Correct Answer MCQ
Which vitamin is fat-soluble and acts as an antioxidant in cell membranes?
Options:
A. Vitamin B1
B. Vitamin C
C. Vitamin E
D. Vitamin B12
Answer & Explanation: Option C. Vitamin E is a fat-soluble vitamin that protects polyunsaturated fatty acids in cell membranes from oxidative damage. Vitamin B1 and B12 are water-soluble vitamins involved in enzymatic reactions. Vitamin C is also water-soluble and functions as an antioxidant in aqueous environments. Vitamin E’s antioxidant property helps maintain membrane integrity, preventing lipid peroxidation, and contributing to cellular defense mechanisms. Its presence in the lipid bilayer ensures protection against reactive oxygen species, supporting healthy cell function.
4. Single Correct Answer MCQ
Which nucleobase is present only in RNA but absent in DNA?
Options:
A. Adenine
B. Cytosine
C. Uracil
D. Guanine
Answer & Explanation: Option C. Uracil is unique to RNA, replacing thymine found in DNA. Adenine, cytosine, and guanine are present in both DNA and RNA. Uracil forms base pairs with adenine during transcription. Its presence allows RNA molecules to function as mRNA, tRNA, and rRNA. Understanding this difference is fundamental in molecular biology and genetics for differentiating DNA from RNA, studying transcription processes, and explaining the flow of genetic information from DNA to protein synthesis.
5. Single Correct Answer MCQ
Which bond is primarily responsible for stabilizing the secondary structure of proteins?
Options:
A. Peptide bond
B. Hydrogen bond
C. Ionic bond
D. Disulfide bond
Answer & Explanation: Option B. Hydrogen bonds stabilize α-helix and β-sheet structures in proteins. Peptide bonds link amino acids linearly, disulfide bonds stabilize tertiary structures, and ionic bonds contribute to tertiary and quaternary interactions. Hydrogen bonding provides directional stability to secondary structures, determining protein folding patterns essential for biological function. Proper secondary structure ensures enzymatic activity, structural support, and interaction with other biomolecules. Understanding hydrogen bonding is crucial in protein chemistry, structural biology, and studying protein misfolding diseases.
6. Single Correct Answer MCQ
Which of the following is an essential fatty acid in humans?
Options:
A. Oleic acid
B. Linoleic acid
C. Palmitic acid
D. Stearic acid
Answer & Explanation: Option B. Linoleic acid is an essential polyunsaturated fatty acid required in the diet, as humans cannot synthesize it. Oleic acid is non-essential, palmitic and stearic acids are saturated fatty acids produced endogenously. Linoleic acid is a precursor for arachidonic acid and eicosanoids, crucial for cellular signaling and inflammation. Recognizing essential fatty acids is important for nutrition, biochemistry, and understanding dietary requirements for maintaining cell membrane integrity and physiological functions.
7. Assertion-Reason MCQ
Assertion (A): RNA contains ribose sugar instead of deoxyribose.
Reason (R): RNA is single-stranded and less stable than DNA.
Options:
A. Both A and R are true, R is the correct explanation of A
B. Both A and R are true, R is not the correct explanation of A
C. A is true, R is false
D. A is false, R is true
Answer & Explanation: Option B. RNA contains ribose sugar making it structurally distinct from DNA which has deoxyribose. RNA is typically single-stranded and less stable due to the 2′-OH group, but this is not the reason for the sugar difference. Understanding sugar composition and strand stability is important in molecular biology, as it affects transcription, translation, and RNA molecule lifespan in the cell.
8. Matching Type MCQ
Match the following enzymes with their substrate:
List-I List-II
(A) Amylase (I) Proteins
(B) Lipase (II) Starch
(C) Protease (III) Lipids
(D) Lactase (IV) Lactose
Options:
1. A-II, B-III, C-I, D-IV
2. A-III, B-II, C-IV, D-I
3. A-IV, B-I, C-III, D-II
4. A-I, B-IV, C-II, D-III
Answer & Explanation: Option 1. Amylase acts on starch (II), Lipase hydrolyzes lipids (III), Protease breaks down proteins (I), and Lactase hydrolyzes lactose (IV). Matching enzyme-substrate pairs is essential in understanding digestion and metabolic processes. Enzymes are highly specific catalysts, and knowing their substrates helps in studying physiological reactions, nutritional science, and clinical enzyme deficiency disorders.
9. Fill in the Blanks / Completion MCQ
The monosaccharide that forms the backbone of nucleotides in DNA is _______.
Options:
A. Ribose
B. Deoxyribose
C. Glucose
D. Fructose
Answer & Explanation: Option B. Deoxyribose is the pentose sugar forming the backbone of DNA nucleotides. Ribose is found in RNA. Glucose and fructose are energy-providing monosaccharides. Deoxyribose contributes to DNA structural stability and enables genetic information storage. Recognizing pentose sugars in nucleotides is crucial for understanding DNA/RNA structure, replication, transcription, and their roles in cellular metabolism.
10. Choose the Correct Statements MCQ
Statement I: Saturated fatty acids have no double bonds and are solid at room temperature.
Statement II: Unsaturated fatty acids contain one or more double bonds and are liquid at room temperature.
Options:
A. Both I and II are correct
B. Only I is correct
C. Only II is correct
D. Both I and II are incorrect
Answer & Explanation: Option A. Saturated fatty acids lack double bonds, resulting in straight chains that pack tightly, making them solid at room temperature. Unsaturated fatty acids contain one or more C=C double bonds, causing kinks that prevent tight packing, making them liquid at room temperature. Understanding fatty acid saturation is essential in nutrition, lipid metabolism, and cellular membrane fluidity regulation.
Topic: Proteins and Hemoglobin Structure; Subtopic: Protein Terminology and Quaternary Structure
Lead Question - 2023
Given below are two statements:
Statement I: A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal).
Statement II: Adult human haemoglobin consists of 4 subunits (two α-type and two β-type).
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is true but Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Answer: (2) Statement I is false but Statement II is true. Proteins have their N-terminal at the start (first amino acid) and C-terminal at the end (last amino acid). Adult human hemoglobin is a tetramer composed of two α-globin and two β-globin subunits. Understanding the correct orientation of protein chains is critical in protein biochemistry, enzyme mechanisms, and structural analysis. Misidentifying the terminals can lead to incorrect interpretation of protein sequences and functionality. Hemoglobin's quaternary structure with α2β2 arrangement is essential for cooperative oxygen binding and efficient transport in blood.
Keyword Definitions:
N-terminal: The start of a protein or polypeptide chain, containing a free amino group.
C-terminal: The end of a protein or polypeptide chain, containing a free carboxyl group.
Haemoglobin: Oxygen-carrying tetrameric protein in red blood cells.
Subunit: Individual polypeptide chain in a multi-chain protein.
α-globin: Alpha subunit of hemoglobin.
β-globin: Beta subunit of hemoglobin.
Quaternary Structure: Arrangement of multiple protein subunits in a multi-subunit complex.
1. Single Correct Answer:
Which terminal of a polypeptide chain has the free amino group?
(1) C-terminal
(2) N-terminal
(3) Both terminals
(4) None
Answer: (2) N-terminal. The N-terminal contains a free amino group and marks the beginning of the polypeptide chain. Correct identification is crucial for sequencing, labeling, and enzymatic modifications.
2. Single Correct Answer:
Adult human hemoglobin has how many subunits?
(1) 2
(2) 3
(3) 4
(4) 5
Answer: (3) 4. Hemoglobin is a tetramer composed of two α-globin and two β-globin subunits, which enables cooperative oxygen binding and efficient transport in blood.
3. Single Correct Answer:
Which globin subunits are present in adult hemoglobin?
(1) α only
(2) β only
(3) α and β
(4) γ only
Answer: (3) α and β. Adult hemoglobin (HbA) consists of two α-globin and two β-globin subunits forming a tetramer necessary for cooperative oxygen binding.
4. Single Correct Answer:
Which structural level describes the arrangement of multiple subunits in hemoglobin?
(1) Primary
(2) Secondary
(3) Tertiary
(4) Quaternary
Answer: (4) Quaternary. Hemoglobin’s quaternary structure refers to its α2β2 tetrameric arrangement, allowing interaction between subunits and cooperative oxygen binding.
5. Single Correct Answer:
Which terminal has the free carboxyl group in a polypeptide?
(1) N-terminal
(2) C-terminal
(3) α-subunit
(4) β-subunit
Answer: (2) C-terminal. The C-terminal is the end of the polypeptide chain with a free carboxyl group, critical in protein chemistry and enzymatic modifications.
6. Single Correct Answer:
Which type of globin is absent in adult hemoglobin?
(1) α
(2) β
(3) γ
(4) Both α and β
Answer: (3) γ. γ-globin is part of fetal hemoglobin (HbF) and is replaced by β-globin in adults. Adult hemoglobin has only α and β subunits.
7. Assertion-Reason:
Assertion A: Hemoglobin is a tetrameric protein.
Reason R: It contains four subunits, two α and two β, enabling cooperative oxygen binding.
(1) A true, R false
(2) A false, R true
(3) Both true, R explains A
(4) Both true, R does not explain A
Answer: (3) Both true, R explains A. Hemoglobin’s tetrameric structure (α2β2) is essential for cooperative oxygen binding, a critical physiological property.
8. Matching Type:
Match the terminus with its property:
List I | List II
A. N-terminal | I. Free carboxyl group
B. C-terminal | II. Free amino group
Options:
(1) A-I, B-II
(2) A-II, B-I
(3) A-I, B-I
(4) A-II, B-II
Answer: (2) A-II, B-I. N-terminal has a free amino group, and C-terminal has a free carboxyl group. Correct understanding is essential for sequencing and protein chemistry studies.
9. Fill in the Blanks:
Adult human hemoglobin consists of two ______ and two ______ subunits.
Options: (1) α, γ (2) α, β (3) β, γ (4) α, α
Answer: (2) α, β. Hemoglobin contains two α-globin and two β-globin chains forming a tetramer for oxygen transport.
10. Choose Correct Statements:
Statement I: N-terminal is the start of the polypeptide chain.
Statement II: C-terminal has a free amino group.
Options:
(1) Both true
(2) I true, II false
(3) I false, II true
(4) Both false
Answer: (2) I true, II false. N-terminal contains a free amino group marking the start, while C-terminal has a free carboxyl group marking the end of the polypeptide chain.
Keyword Definitions:
Enzyme: Biological catalyst made of proteins that speeds up biochemical reactions without being consumed.
Denaturation: Structural alteration of proteins by heat or chemicals leading to loss of biological activity.
Competitive inhibitor: Molecule resembling the substrate that competes for the enzyme’s active site and inhibits its action.
Substrate: The specific molecule upon which an enzyme acts to form products during catalysis.
Active site: Region of an enzyme where the substrate binds and undergoes a chemical reaction.
Temperature effect: Low temperature slows enzyme activity while high temperature causes denaturation and permanent inactivation.
Enzyme inhibition: Process of decreasing enzyme activity by specific molecules or environmental factors.
Lock and key model: Concept explaining enzyme specificity where substrate fits precisely into enzyme’s active site.
Activation energy: Minimum energy required for a chemical reaction to occur; enzymes lower this energy barrier.
Cofactor: Non-protein component required for some enzymes to exhibit full catalytic activity.
Optimum temperature: Specific temperature range where an enzyme shows maximum activity and stability.
Lead Question - 2023:
Given below are two statements:
Statement I: Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
Statement II: When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor.
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is true but Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Answer & Explanation: (3) Both Statement I and Statement II are true. Low temperature slows down molecular motion, temporarily inactivating enzymes but preserving their structure. High temperature causes irreversible denaturation of proteins, destroying enzyme function. A competitive inhibitor mimics the substrate structure and binds to the enzyme’s active site, blocking substrate access. Both mechanisms illustrate enzyme sensitivity and regulatory control in biochemical systems.
1. Which of the following statements about enzyme activity is correct?
(1) Enzymes work faster at any temperature
(2) High temperature always increases enzyme efficiency
(3) Enzyme activity decreases at very high temperatures due to denaturation
(4) Enzyme activity remains constant irrespective of pH
Explanation: Enzyme activity rises with temperature up to an optimum level, after which denaturation occurs, leading to permanent loss of activity. Correct answer is (3). High heat breaks hydrogen bonds and disrupts the active site, rendering the enzyme non-functional and unable to catalyze reactions effectively in living cells.
2. Which one acts as a competitive inhibitor for succinic dehydrogenase?
(1) Malonic acid
(2) Fumaric acid
(3) Oxaloacetic acid
(4) Acetic acid
Explanation: Malonic acid closely resembles the structure of succinic acid and competes for the enzyme succinic dehydrogenase’s active site. Correct answer is (1). It prevents substrate binding, reducing product formation. This classical example demonstrates the mechanism of competitive inhibition regulating metabolic reactions reversibly and specifically.
3. Enzymes are destroyed at high temperatures because:
(1) They evaporate
(2) They coagulate and lose structure
(3) Substrate gets destroyed
(4) They convert to cofactors
Explanation: Enzymes lose their active three-dimensional structure at high temperatures due to denaturation of protein molecules. Correct answer is (2). This process disrupts the peptide bonds and alters the active site, preventing substrate binding, resulting in loss of catalytic function permanently in most cases.
4. The rate of enzyme reaction is maximum at:
(1) 0°C
(2) 25°C
(3) Optimum temperature
(4) 90°C
Explanation: Enzyme activity is highest at an optimum temperature where molecular collisions and active site conformation are ideal. Correct answer is (3). Deviation above or below this point reduces reaction rate either by slowing molecular motion or by denaturing the enzyme’s protein structure, affecting catalysis efficiency.
5. Enzyme inhibition by a molecule similar to substrate is called:
(1) Allosteric inhibition
(2) Non-competitive inhibition
(3) Competitive inhibition
(4) Irreversible inhibition
Explanation: In competitive inhibition, inhibitor molecules structurally resemble the substrate and compete for the enzyme’s active site. Correct answer is (3). This reduces enzyme-substrate complex formation, thereby decreasing reaction rate. The effect can be reversed by increasing substrate concentration, restoring enzymatic activity efficiently.
6. Which of the following factors does not affect enzyme activity?
(1) Temperature
(2) pH
(3) Light intensity
(4) Substrate concentration
Explanation: Enzyme activity depends on temperature, pH, and substrate concentration but not on light intensity. Correct answer is (3). Changes in temperature and pH alter enzyme structure, while substrate availability affects binding frequency. Light affects photosynthesis but not general enzyme-mediated biochemical reactions directly in cellular metabolism.
Assertion-Reason Question
7. Assertion (A): Competitive inhibition can be overcome by increasing substrate concentration.
Reason (R): Substrate and inhibitor compete for the same active site of the enzyme.
(1) Both A and R true, and R explains A
(2) Both A and R true, but R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Both statements are true and R correctly explains A. In competitive inhibition, increasing substrate concentration outcompetes the inhibitor, restoring enzyme activity. Correct answer is (1). This reversible inhibition regulates metabolic pathways efficiently without permanently damaging enzyme structure or function.
Matching Type Question
8. Match the enzyme with its function:
A. Sucrase – i. Converts sucrose to glucose and fructose
B. Lipase – ii. Hydrolyses fats into fatty acids and glycerol
C. Urease – iii. Breaks down urea into ammonia and CO2
D. Amylase – iv. Converts starch to maltose
(1) A-i, B-ii, C-iii, D-iv
(2) A-ii, B-i, C-iv, D-iii
(3) A-iv, B-iii, C-ii, D-i
(4) A-i, B-iv, C-ii, D-iii
Explanation: Each enzyme performs a specific catalytic role. Sucrase splits sucrose, lipase hydrolyzes fats, urease decomposes urea, and amylase digests starch. Correct answer is (1). Enzyme specificity arises from structural complementarity between the active site and substrate, ensuring precise and regulated biochemical reactions.
Fill in the Blanks Question
9. At low temperatures, enzyme activity _______ due to decreased molecular motion.
(1) Increases
(2) Decreases
(3) Remains constant
(4) Becomes irreversible
Explanation: At low temperatures, enzymes become temporarily inactive as molecular collisions and substrate binding slow down. Correct answer is (2). However, the enzyme structure remains intact, and activity can resume when temperature returns to the optimum range, showing reversible inactivation rather than denaturation.
Choose the Correct Statements Question
10. Statement I: Enzymes increase the activation energy of a reaction.
Statement II: Competitive inhibitors resemble substrates and compete for the active site.
(1) Both I and II true
(2) Only I true
(3) Only II true
(4) Both false
Explanation: Enzymes lower, not increase, activation energy to accelerate reactions. Competitive inhibitors structurally mimic substrates, blocking access to the enzyme’s active site. Correct answer is (3). Thus, enzymes act as catalysts, and inhibitors regulate their efficiency through reversible competition mechanisms in metabolism.
Keyword Definitions:
Melonate: A compound with antimicrobial properties that inhibits specific bacterial enzymes.
Pathogenic bacteria: Bacteria capable of causing disease in host organisms.
Enzyme inhibition: A process where a molecule decreases or stops enzyme activity.
Lipase: Enzyme that breaks down lipids into glycerol and fatty acids.
Dinitrogenase: Nitrogen-fixing enzyme responsible for converting atmospheric nitrogen to ammonia.
Succinic dehydrogenase: Enzyme of the citric acid cycle, catalyzing succinate to fumarate oxidation.
Amylase: Enzyme that breaks down starch into simple sugars.
Growth inhibition: Reduction or cessation of microbial proliferation.
Antimicrobial: Substance that kills or inhibits microorganisms.
Mechanism of action: The specific biochemical interaction through which a compound affects a microorganism.
Pathogenicity: The ability of an organism to cause disease.
Lead Question - 2023:
Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of:
(1) Lipase
(2) Dinitrogenase
(3) Succinic dehydrogenase
(4) Amylase
Answer & Explanation: (3) Succinic dehydrogenase. Melonate targets succinic dehydrogenase, an essential enzyme of the bacterial citric acid cycle, disrupting energy metabolism. Inhibition of this enzyme prevents conversion of succinate to fumarate, reducing ATP production and ultimately suppressing bacterial growth. Unlike lipase, dinitrogenase, or amylase, succinic dehydrogenase is critical for the survival of many pathogenic bacteria. Therefore, by targeting this enzyme, melonate acts as an effective antimicrobial agent, ensuring decreased proliferation and survival of harmful bacterial species in infected environments.
1. Which enzyme is crucial for the citric acid cycle in bacteria?
(1) Lipase
(2) Succinic dehydrogenase
(3) Amylase
(4) Dinitrogenase
Explanation: Succinic dehydrogenase catalyzes succinate to fumarate in the citric acid cycle, essential for energy production in bacteria. Correct answer is (2).
2. Melonate primarily inhibits bacterial growth by targeting:
(1) Lipid metabolism
(2) Carbohydrate metabolism
(3) Citric acid cycle
(4) Nitrogen fixation
Explanation: Melonate inhibits succinic dehydrogenase, disrupting the citric acid cycle and ATP production, leading to bacterial growth inhibition. Correct answer is (3).
3. Inhibition of which enzyme reduces ATP production in bacteria?
(1) Dinitrogenase
(2) Succinic dehydrogenase
(3) Amylase
(4) Lipase
Explanation: Succinic dehydrogenase is part of the citric acid cycle; its inhibition lowers ATP synthesis and energy availability, limiting bacterial proliferation. Correct answer is (2).
4. Which enzyme is not a target of melonate?
(1) Succinic dehydrogenase
(2) Lipase
(3) Both 1 and 2
(4) None
Explanation: Melonate specifically inhibits succinic dehydrogenase; it does not inhibit lipase, dinitrogenase, or amylase. Correct answer is (2).
5. Which process is directly affected by succinic dehydrogenase inhibition?
(1) Nitrogen fixation
(2) ATP synthesis
(3) Starch breakdown
(4) Lipid hydrolysis
Explanation: Succinic dehydrogenase participates in the citric acid cycle, essential for ATP synthesis. Inhibition reduces energy supply to bacteria. Correct answer is (2).
6. Which enzyme is involved in converting succinate to fumarate?
(1) Lipase
(2) Succinic dehydrogenase
(3) Dinitrogenase
(4) Amylase
Explanation: Succinic dehydrogenase catalyzes succinate oxidation to fumarate in the citric acid cycle, crucial for energy metabolism. Correct answer is (2).
Assertion-Reason Type Question
7. Assertion (A): Melonate inhibits bacterial energy metabolism.
Reason (R): Succinic dehydrogenase is an essential enzyme in the citric acid cycle.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Inhibiting succinic dehydrogenase disrupts the citric acid cycle, reducing ATP production and bacterial energy metabolism. Both statements are true, and R explains A. Correct answer is (1).
Matching Type Question
8. Match the enzyme with its primary function:
A. Succinic dehydrogenase – i. Citric acid cycle
B. Lipase – ii. Lipid breakdown
C. Amylase – iii. Starch hydrolysis
D. Dinitrogenase – iv. Nitrogen fixation
(1) A-i, B-ii, C-iii, D-iv
(2) A-ii, B-i, C-iv, D-iii
(3) A-i, B-iii, C-ii, D-iv
(4) A-iv, B-iii, C-i, D-ii
Explanation: Succinic dehydrogenase functions in citric acid cycle, lipase in lipid breakdown, amylase in starch hydrolysis, and dinitrogenase in nitrogen fixation. Correct answer is (1).
Fill in the Blanks Question
9. Melonate inhibits bacterial growth by targeting _______ in the citric acid cycle.
(1) Lipase
(2) Succinic dehydrogenase
(3) Amylase
(4) Dinitrogenase
Explanation: Succinic dehydrogenase catalyzes succinate to fumarate; its inhibition by melonate decreases ATP synthesis, reducing bacterial growth. Correct answer is (2).
Choose the Correct Statements Question
10. Statement I: Succinic dehydrogenase is part of the citric acid cycle.
Statement II: Melonate inhibits bacterial growth by blocking lipid breakdown.
(1) Both I and II correct
(2) Only I correct
(3) Only II correct
(4) Neither I nor II correct
Explanation: Succinic dehydrogenase is a key citric acid cycle enzyme (Statement I correct). Melonate targets succinic dehydrogenase, not lipid metabolism (Statement II incorrect). Correct answer is (2).
Keyword Definitions:
Cellulose: A structural polysaccharide composed of β-D-glucose units linked by β-1,4-glycosidic bonds forming linear chains.
Iodine Test: A chemical test where iodine interacts with polysaccharides to produce color depending on structure.
Helical Structure: Spiral conformation that can trap iodine molecules, producing characteristic colors.
Disaccharide: Carbohydrate formed by two monosaccharide units, e.g., sucrose, maltose.
Polysaccharide: Long-chain carbohydrate composed of multiple monosaccharide units.
Starch: Polysaccharide of glucose forming helical structures; gives blue color with iodine.
β-Glucose: Glucose in beta configuration; forms straight cellulose chains.
Color Reaction: Visual indication in chemical tests due to molecular interactions.
Structural Polysaccharide: Provides mechanical strength, e.g., cellulose in plant cell walls.
α-Glucose: Glucose in alpha configuration; forms coiled starch molecules.
Complex Helices: Spiral structures in polysaccharides that can trap iodine molecules for color formation.
Lead Question - 2023:
Cellulose does not form blue colour with Iodine because:
(1) It does not contain complex helices and hence cannot hold iodine molecules
(2) It breaks down when iodine reacts with it
(3) It is a disaccharide
(4) It is a helical molecule
Answer & Explanation: (1) It does not contain complex helices and hence cannot hold iodine molecules. Cellulose is a linear polysaccharide composed of β-D-glucose units linked by β-1,4-glycosidic bonds forming straight chains. Unlike starch, which forms α-helices capable of trapping iodine molecules and producing a blue color, cellulose chains do not coil into helices. Therefore, iodine cannot be trapped, and no blue coloration occurs. The β-configuration prevents helix formation. Cellulose’s function is structural, providing rigidity to plant cell walls, unlike starch, which serves as energy storage and reacts with iodine to produce a characteristic blue color in qualitative tests.
1. Starch gives which color with iodine?
(1) Blue
(2) Red
(3) Brown
(4) Green
Explanation: Starch has α-helical structure that traps iodine molecules producing blue color. Correct answer is (1).
2. Cellulose is made of which glucose type?
(1) α-D-glucose
(2) β-D-glucose
(3) γ-D-glucose
(4) δ-D-glucose
Explanation: Cellulose is composed of β-D-glucose units forming linear chains. Correct answer is (2).
3. Which polysaccharide is primarily for energy storage?
(1) Cellulose
(2) Starch
(3) Chitin
(4) Pectin
Explanation: Starch stores energy in plants, while cellulose provides structural support. Correct answer is (2).
4. Chitin is a polysaccharide composed of:
(1) Glucose
(2) N-acetylglucosamine
(3) Fructose
(4) Sucrose
Explanation: Chitin is made of N-acetylglucosamine units forming structural exoskeletons in arthropods. Correct answer is (2).
5. Which polysaccharide has β-1,4 linkages?
(1) Starch
(2) Cellulose
(3) Glycogen
(4) Maltose
Explanation: Cellulose has β-1,4-glycosidic bonds forming linear chains. Correct answer is (2).
6. Glycogen is mainly stored in:
(1) Plant roots
(2) Animal liver and muscles
(3) Cell walls
(4) Chloroplasts
Explanation: Glycogen is energy storage polysaccharide in liver and muscles. Correct answer is (2).
Assertion-Reason Type Question
7. Assertion (A): Starch gives blue color with iodine.
Reason (R): Starch molecules form α-helices that trap iodine.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Starch α-helices trap iodine molecules giving blue color; both statements are correct and R explains A. Correct answer is (1).
Matching Type Question
8. Match polysaccharide with main function:
A. Cellulose – (i) Energy storage
B. Starch – (ii) Structural support
C. Glycogen – (iii) Energy storage in animals
D. Chitin – (iv) Exoskeleton
(1) A-(ii), B-(i), C-(iii), D-(iv)
(2) A-(i), B-(ii), C-(iii), D-(iv)
(3) A-(iv), B-(iii), C-(ii), D-(i)
(4) A-(ii), B-(iv), C-(i), D-(iii)
Explanation: Correct functional matches: Cellulose-structural; Starch-energy; Glycogen-animal energy; Chitin-exoskeleton. Correct answer is (1).
Fill in the Blanks Question
9. The polysaccharide that does not react with iodine to give blue color is ______.
(1) Starch
(2) Glycogen
(3) Cellulose
(4) Maltose
Explanation: Cellulose’s linear chains cannot form helices to trap iodine; no blue color occurs. Correct answer is (3).
Choose the Correct Statements Question
10. Statement I: Starch forms α-helical structures.
Statement II: Cellulose forms β-linear chains.
(1) I true, II false
(2) I false, II true
(3) Both true
(4) Both false
Explanation: Starch forms α-helices; cellulose forms linear β-chains. Both statements are correct. Correct answer is (3).
Chapter : Biomolecules; Topic : Enzymes; Subtopic : Catalase and Prosthetic Groups
Keyword Definitions :
Enzyme : Biological catalyst that speeds up biochemical reactions without being consumed.
Prosthetic group : A non-protein component tightly bound to an enzyme and essential for its activity.
Catalase : An enzyme that catalyzes the breakdown of hydrogen peroxide into water and oxygen.
Hydrogen peroxide (H2O2) : A toxic byproduct of metabolism that must be decomposed rapidly in cells.
Haem : Iron-containing porphyrin group serving as a prosthetic group in enzymes like catalase and peroxidase.
Oxidoreductase : A class of enzymes involved in oxidation-reduction reactions.
Lead Question - 2022 (Ganganagar)
In the enzyme which catalyses the breakdown of:
H2O2 → H2O + O2
the prosthetic group is :
1. Nicotinamide adenine dinucleotide
2. Haem
3. Zinc
4. Niacin
Explanation : Catalase is the enzyme responsible for breaking down hydrogen peroxide into water and oxygen, preventing oxidative damage to cells. Its prosthetic group is Haem, an iron-porphyrin complex that facilitates the decomposition of H2O2. The iron center alternates between oxidation states, allowing rapid reaction turnover. Thus, Option 2 is correct.
1. The enzyme catalase is mainly present in which cellular organelle?
1. Ribosome
2. Peroxisome
3. Lysosome
4. Mitochondria
Explanation : Catalase is localized in peroxisomes, where it decomposes hydrogen peroxide generated during fatty acid oxidation and other metabolic activities. This prevents cellular toxicity. Peroxisomes serve as detoxification centers, protecting cells from oxidative stress. Hence, Option 2 is correct.
2. The enzyme which catalyzes the conversion of hydrogen peroxide into water and oxygen belongs to which enzyme class?
1. Hydrolase
2. Oxidoreductase
3. Transferase
4. Lyase
Explanation : Catalase belongs to the Oxidoreductase class of enzymes, which catalyze oxidation-reduction reactions by transferring electrons from donor to acceptor molecules. Catalase oxidizes hydrogen peroxide while reducing another molecule of the same compound. Therefore, Option 2 is correct.
3. Which of the following enzymes contains zinc as its prosthetic group?
1. Carbonic anhydrase
2. Catalase
3. Peroxidase
4. Cytochrome oxidase
Explanation : Zinc acts as a prosthetic group in carbonic anhydrase, facilitating the reversible hydration of carbon dioxide to carbonic acid. This reaction is crucial for CO2 transport and pH regulation. Catalase, however, contains haem instead of zinc. Thus, Option 1 is correct.
4. Which of the following statements about catalase is correct?
1. It is a zinc-containing enzyme.
2. It decomposes H2O2 into water and oxygen.
3. It is inactive in the presence of iron.
4. It is found only in prokaryotes.
Explanation : Catalase catalyzes the decomposition of hydrogen peroxide into water and oxygen, protecting cells from oxidative injury. It contains haem as its prosthetic group, with iron as the reactive center. It is found in almost all aerobic organisms. Hence, Option 2 is correct.
5. Assertion (A): Catalase protects cells from oxidative damage.
Reason (R): Catalase converts toxic hydrogen peroxide into harmless products.
1. Both A and R are true, and R is the correct explanation of A
2. Both A and R are true, but R is not the correct explanation of A
3. A is true, R is false
4. A is false, R is true
Explanation : Hydrogen peroxide is toxic to cells as it forms free radicals. Catalase detoxifies H2O2 into water and oxygen, preventing oxidative injury. Hence, both statements are true, and R correctly explains A. Therefore, Option 1 is correct.
6. Which vitamin-derived coenzyme participates in dehydrogenase reactions but not in catalase action?
1. FAD
2. NAD+
3. NADP+
4. Biotin
Explanation : NAD+ (Nicotinamide adenine dinucleotide) is a coenzyme derived from niacin (Vitamin B3) and participates in various dehydrogenase reactions by transferring electrons. Catalase does not require NAD+, as its prosthetic group is haem. Thus, Option 2 is correct.
7. (Matching Type)
Match List-I with List-II:
List-I List-II
(a) Catalase (i) Haem
(b) Carbonic anhydrase (ii) Zn2+
(c) Urease (iii) Ni2+
(d) Nitrogenase (iv) Mo
1. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
2. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
3. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
4. (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
Explanation : Catalase contains haem (iron), carbonic anhydrase has zinc, urease has nickel, and nitrogenase contains molybdenum. These metals are essential cofactors for their catalytic activity. Therefore, Option 1 is correct.
8. (Fill in the Blank)
Catalase enzyme is classified under the category of _______ enzymes.
1. Oxidoreductase
2. Hydrolase
3. Isomerase
4. Ligase
Explanation : Catalase belongs to the Oxidoreductase class of enzymes because it catalyzes redox reactions, facilitating the decomposition of hydrogen peroxide by oxidation-reduction. The enzyme’s haem group enables the transfer of electrons during the reaction. Thus, Option 1 is correct.
9. Which of the following statements about prosthetic groups is incorrect?
1. They are loosely bound to enzymes.
2. They are essential for enzyme activity.
3. They can be metal ions or organic molecules.
4. They remain attached during catalysis.
Explanation : Prosthetic groups are tightly bound, non-protein components of enzymes, unlike coenzymes which are loosely bound. They are essential for catalytic function and may be organic molecules or metal ions. Hence, the incorrect statement is that they are loosely bound. Therefore, Option 1 is correct.
10. (Choose the correct statements)
Statement I: Catalase contains a haem prosthetic group.
Statement II: Catalase catalyzes the oxidation of H2O2 into water and oxygen.
1. Both statements are correct
2. Only Statement I is correct
3. Only Statement II is correct
4. Both statements are incorrect
Explanation : Catalase is a haem-containing enzyme that decomposes hydrogen peroxide into water and oxygen, protecting cells from oxidative stress. Both statements accurately describe the enzyme’s structure and function, making Option 1 correct.
Chapter: Biomolecules; Topic: Amino Acids and Proteins; Subtopic: Structure and Properties of Amino Acids
Keyword Definitions:
• Amino acids: Organic compounds containing both amino (-NH₂) and carboxyl (-COOH) groups.
• Ionizable groups: Functional groups that can donate or accept protons, changing charge with pH.
• Zwitterion: Molecule with both positive and negative charges but overall neutral.
• pH: Measure of hydrogen ion concentration; acidic pH 7.
• Isoelectric point (pI): pH at which amino acid exists predominantly as zwitterion.
• Acidic pH: Environment with high H⁺ concentration.
• Basic pH: Environment with low H⁺ concentration.
• Protein structure: Determined by amino acid sequence and side chain ionization.
• Buffering: Ability of amino acids to resist changes in pH.
• Charge separation: Existence of positive and negative charges on the same molecule.
• Electrophoretic mobility: Movement of charged amino acids in an electric field depending on pH.
Lead Question - 2022 (Ganganagar)
Given below are two statements:
Statement I: Amino acids have a property of ionizable nature of –NH2 and –COOH groups, hence have different structures at different pH.
Statement II: Amino acids can exist as Zwitterionic form at acidic and basic pH.
In the light of the above statements, choose the most appropriate answer from the options given below:
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are Incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Explanation: Statement I is correct as amino acids have ionizable –NH2 and –COOH groups which can gain or lose protons depending on pH, leading to different structural forms. Statement II is incorrect because amino acids exist as zwitterions predominantly near their isoelectric point (neutral pH), not at strongly acidic or basic pH where they carry net positive or negative charge. Hence, only Statement I is correct while Statement II is wrong. Correct answer is 3.
1. Single Correct Answer MCQ:
At physiological pH (~7.4), an amino acid typically exists as:
a) Fully protonated cation
b) Fully deprotonated anion
c) Zwitterion
d) Neutral molecule with no charges
Explanation: At physiological pH, the amino group is protonated (+) and the carboxyl group is deprotonated (–), forming a zwitterion with no net charge. Correct answer is c.
2. Single Correct Answer MCQ:
The isoelectric point (pI) of an amino acid is:
a) pH at which amino acid is fully protonated
b) pH at which amino acid exists predominantly as zwitterion
c) pH at which amino acid is fully deprotonated
d) pH at which amino acid denatures
Explanation: pI is the pH where amino acid has equal positive and negative charges (zwitterion), overall neutral. Correct answer is b.
3. Single Correct Answer MCQ:
Which groups in amino acids are ionizable?
a) Only –NH2
b) Only –COOH
c) –NH2 and –COOH
d) Side chains only
Explanation: Both the amino (-NH2) and carboxyl (-COOH) groups are ionizable and determine amino acid charge at different pH. Correct answer is c.
4. Single Correct Answer MCQ:
At low pH, amino acids mainly exist as:
a) Cations
b) Anions
c) Zwitterions
d) Neutral molecules
Explanation: In acidic environment, –NH2 is protonated (+) and –COOH remains protonated, giving net positive charge (cation). Correct answer is a.
5. Single Correct Answer MCQ:
At high pH, amino acids predominantly exist as:
a) Cations
b) Anions
c) Zwitterions
d) Neutral molecules
Explanation: In basic environment, amino group is deprotonated and carboxyl group is deprotonated, giving net negative charge (anion). Correct answer is b.
6. Single Correct Answer MCQ:
Zwitterionic form of amino acids occurs due to:
a) Complete protonation of both groups
b) Deprotonation of both groups
c) Protonated amino group and deprotonated carboxyl group
d) Side chain ionization only
Explanation: Zwitterion arises when amino group is protonated (+) and carboxyl group is deprotonated (–), yielding no net charge. Correct answer is c.
7. Assertion-Reason MCQ:
Assertion (A): Amino acids change form at different pH.
Reason (R): –NH2 and –COOH groups ionize at different pH values.
a) Both A and R true, R explains A
b) Both A and R true, R does not explain A
c) A true, R false
d) Both false
Explanation: Ionizable groups respond to pH, causing structural and charge changes. Both A and R are true and R explains A. Correct answer is a.
8. Matching Type MCQ:
Match List-I with List-II:
List-I | List-II
(a) Acidic pH | (i) Cation
(b) Neutral pH | (ii) Zwitterion
(c) Basic pH | (iii) Anion
Options:
1. a-i, b-ii, c-iii
2. a-ii, b-i, c-iii
3. a-iii, b-ii, c-i
4. a-i, b-iii, c-ii
Explanation: Acidic pH gives net positive (cation), neutral pH is zwitterion, and basic pH gives net negative (anion). Correct answer is 1.
9. Fill in the Blanks / Completion MCQ:
At isoelectric point, amino acid has ________ net charge.
a) Positive
b) Negative
c) Zero
d) Depends on side chain
Explanation: At pI, the amino acid exists as zwitterion with net zero charge. Correct answer is c.
10. Choose the correct statements MCQ (Statement I & II):
Statement I: Amino acids carry different charges at different pH.
Statement II: Zwitterionic form occurs at extreme acidic pH.
a) Both I and II correct
b) Only I correct
c) Only II correct
d) Both incorrect
Explanation: Statement I is correct; amino acids ionize with pH changes. Statement II is incorrect; zwitterionic form occurs near neutral pH, not extreme acidic. Correct answer is b.
Chapter: Biomolecules and Enzymes; Topic: Carbohydrate and Protein Metabolism; Subtopic: Enzymatic Reactions and Specificity
Keyword Definitions:
• Enzyme: Biological catalyst that speeds up chemical reactions without being consumed.
• Maltose: Disaccharide composed of two glucose units.
• Maltase: Enzyme that hydrolyzes maltose into glucose molecules.
• Sucrose: Disaccharide of glucose and fructose.
• Sucrase: Enzyme that converts sucrose into glucose and fructose.
• Lactose: Disaccharide composed of glucose and galactose.
• Lactase: Enzyme that hydrolyzes lactose into glucose and galactose.
• Dipeptides: Molecules consisting of two amino acids linked by a peptide bond.
• Peptidases: Enzymes that hydrolyze peptide bonds into amino acids.
• Hydrolysis: Reaction involving the breakdown of a compound by water.
• Specificity: Enzyme acts only on a particular substrate or reaction.
Lead Question - 2022 (Ganganagar)
Choose the incorrect enzymatic reaction:
Maltose → Maltase → Glucose + Galactose
Sucrose → Sucrase → Glucose + Fructose
Lactose → Lactase → Glucose + Galactose
Dipeptides → Dipeptidases → Amino acids
Explanation: Maltose consists of two glucose molecules, not glucose and galactose. Therefore, the reaction Maltose → Maltase → Glucose + Galactose is incorrect. Correct reaction: Maltose → Maltase → 2 Glucose. The other enzymatic reactions are accurate: sucrose hydrolyzed by sucrase yields glucose and fructose, lactose hydrolyzed by lactase yields glucose and galactose, and dipeptides hydrolyzed by peptidases yield amino acids. Enzyme specificity ensures each enzyme acts on its proper substrate, and misrepresenting substrates leads to an incorrect reaction. Correct answer is the first reaction.
1. Single Correct Answer MCQ:
Which enzyme hydrolyzes lactose into glucose and galactose?
a) Maltase
b) Lactase
c) Sucrase
d) Peptidase
Explanation: Lactase specifically hydrolyzes lactose into glucose and galactose. Maltase acts on maltose, sucrase on sucrose, and peptidases on peptide bonds. Correct answer is b.
2. Single Correct Answer MCQ:
Maltose is composed of:
a) Glucose + Fructose
b) Glucose + Galactose
c) 2 Glucose molecules
d) Fructose + Galactose
Explanation: Maltose is a disaccharide of two glucose molecules, which is hydrolyzed by maltase. Correct answer is c.
3. Single Correct Answer MCQ:
Sucrase catalyzes the hydrolysis of:
a) Maltose
b) Lactose
c) Sucrose
d) Dipeptides
Explanation: Sucrase hydrolyzes sucrose into glucose and fructose. Correct answer is c.
4. Single Correct Answer MCQ:
Which enzyme breaks peptide bonds in dipeptides?
a) Maltase
b) Lactase
c) Sucrase
d) Peptidases
Explanation: Peptidases hydrolyze peptide bonds to release amino acids. Correct answer is d.
5. Single Correct Answer MCQ:
Which of the following is an incorrect enzyme-substrate pair?
a) Maltose - Maltase
b) Lactose - Lactase
c) Sucrose - Sucrase
d) Maltose - Lactase
Explanation: Maltose is hydrolyzed by maltase, not lactase. Correct answer is d.
6. Single Correct Answer MCQ:
The product of sucrose hydrolysis by sucrase is:
a) 2 Glucose
b) Glucose + Fructose
c) Glucose + Galactose
d) Amino acids
Explanation: Sucrase breaks sucrose into glucose and fructose. Correct answer is b.
7. Assertion-Reason MCQ:
Assertion (A): Maltose hydrolysis produces glucose + galactose.
Reason (R): Maltose consists of glucose and galactose.
a) Both A and R true, R explains A
b) Both A and R true, R does not explain A
c) A true, R false
d) Both false
Explanation: Both assertion and reason are false; maltose yields 2 glucose molecules. Correct answer is d.
8. Matching Type MCQ:
Match List-I with List-II:
List-I | List-II
(a) Lactose | (i) Glucose + Fructose
(b) Sucrose | (ii) 2 Glucose
(c) Maltose | (iii) Glucose + Galactose
(d) Dipeptides | (iv) Amino acids
Options:
1. a-iii, b-i, c-ii, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-iv, b-i, c-ii, d-iii
Explanation: Correct matches are: Lactose (iii), Sucrose (i), Maltose (ii), Dipeptides (iv). Correct answer is 1.
9. Fill in the Blanks / Completion MCQ:
Maltose hydrolyzed by maltase produces _______.
a) Glucose + Fructose
b) 2 Glucose molecules
c) Glucose + Galactose
d) Amino acids
Explanation: Maltose consists of two glucose units. Maltase hydrolyzes it into 2 glucose molecules. Correct answer is b.
10. Choose the correct statements MCQ (Statement I & II):
Statement I: Lactose is hydrolyzed by lactase into glucose and galactose.
Statement II: Maltose hydrolyzed by maltase produces glucose + galactose.
a) Both I and II correct
b) Only I correct
c) Only II correct
d) Both incorrect
Explanation: Statement I is correct. Statement II is incorrect because maltose yields 2 glucose molecules. Correct answer is b.
Chapter: Biomolecules; Topic: Proteins; Subtopic: Structure of Proteins
Keyword Definitions:
• Primary protein: Protein with a specific sequence of amino acids linked linearly by peptide bonds.
• Polypeptide: A linear polymer of amino acids joined by peptide bonds forming a protein chain.
• Peptide bond: Covalent bond linking carboxyl group of one amino acid to amino group of another.
• Amino acids: Building blocks of proteins containing amino (-NH₂) and carboxyl (-COOH) groups.
• Linear chain: Sequence of amino acids arranged in a straight chain without folding.
• Conformation: Three-dimensional shape adopted by proteins due to folding.
• Polymer: Molecule composed of repeating monomeric units.
• Secondary structure: Alpha helices and beta sheets formed by hydrogen bonding in protein.
• Tertiary structure: Three-dimensional folding of a single polypeptide chain.
• Quaternary structure: Arrangement of multiple polypeptide chains into a functional protein complex.
• Protein function: Determined by sequence, structure, and folding of amino acids.
Lead Question - 2022 (Ganganagar)
Primary proteins are also called as polypeptides because:
1. They are linear chains
2. They are polymers of peptide monomers
3. Successive amino acids are joined by peptide bonds
4. They can assume many conformations
Explanation: Primary proteins consist of linear sequences of amino acids linked by peptide bonds, forming a chain called a polypeptide. They are polymers where each monomer is an amino acid. While proteins can fold into secondary, tertiary, and quaternary structures, the primary structure strictly refers to the sequence of amino acids. The peptide bond forms the backbone of this linear polymer. Correct answer is 3. Successive amino acids are joined by peptide bonds. This covalent linkage defines the polypeptide nature of primary proteins and determines the sequence, which ultimately dictates higher-level folding and the protein’s functional properties.
1. Single Correct Answer MCQ:
Which bond links amino acids in a polypeptide chain?
a) Hydrogen bond
b) Peptide bond
c) Disulfide bond
d) Ionic bond
Explanation: Amino acids in a polypeptide are joined by covalent peptide bonds formed between carboxyl and amino groups. Hydrogen, disulfide, and ionic bonds contribute to higher-level folding. Correct answer is b) Peptide bond.
2. Single Correct Answer MCQ:
Primary structure of protein determines:
a) 3D folding
b) Amino acid sequence
c) Quaternary structure
d) Denaturation rate
Explanation: The primary structure is the specific linear sequence of amino acids. This sequence dictates folding and higher structures but itself is simply the chain of residues. Correct answer is b) Amino acid sequence.
3. Single Correct Answer MCQ:
Polypeptide chain is:
a) Branched polymer
b) Linear polymer of amino acids
c) Single amino acid
d) Disaccharide chain
Explanation: A polypeptide is a linear polymer of amino acids linked by peptide bonds, forming the backbone of proteins. Correct answer is b) Linear polymer of amino acids.
4. Single Correct Answer MCQ:
Which structure refers only to sequence of amino acids?
a) Primary
b) Secondary
c) Tertiary
d) Quaternary
Explanation: Primary structure is the linear amino acid sequence connected by peptide bonds. Secondary and higher structures arise from folding. Correct answer is a) Primary.
5. Single Correct Answer MCQ:
Disulfide bridges in proteins are part of:
a) Primary structure
b) Secondary structure
c) Tertiary structure
d) None of the above
Explanation: Disulfide bonds form between cysteine residues, stabilizing tertiary structure, not primary. Correct answer is c) Tertiary structure.
6. Single Correct Answer MCQ:
Which is true about peptide bonds?
a) Form between R-groups
b) Covalent bond linking amino acids
c) Hydrogen bonds holding helices
d) Weak ionic interactions
Explanation: Peptide bonds are covalent linkages between carboxyl and amino groups of amino acids forming polypeptides. Correct answer is b) Covalent bond linking amino acids.
7. Assertion-Reason MCQ:
Assertion (A): Proteins are called polypeptides.
Reason (R): They are polymers of amino acids joined by peptide bonds.
a) Both A and R are true, R explains A
b) Both A and R are true, R does not explain A
c) A is true, R is false
d) A is false, R is true
Explanation: Proteins consist of linear chains of amino acids linked by peptide bonds. This is why primary proteins are called polypeptides. Both assertion and reason are true, with reason explaining the assertion. Correct answer is a) Both A and R are true, R explains A.
8. Matching Type MCQ:
Match structures with descriptions:
Column I
A) Primary
B) Secondary
C) Tertiary
D) Quaternary
Column II
1) Linear amino acid sequence
2) Alpha helices and beta sheets
3) 3D folding of single chain
4) Assembly of multiple chains
Choices:
A-__ B-__ C-__ D-__
Explanation: Correct matches: A-1 (primary – sequence), B-2 (secondary – helices/sheets), C-3 (tertiary – 3D folding), D-4 (quaternary – multiple chains). Primary is the polypeptide chain.
9. Fill in the Blanks / Completion MCQ:
The sequence of amino acids joined by peptide bonds is called ______.
a) Disaccharide
b) Polypeptide
c) Nucleotide
d) Lipid
Explanation: A polypeptide is a linear sequence of amino acids linked via peptide bonds, forming the primary structure of proteins. Correct answer is b) Polypeptide.
10. Choose the correct statements MCQ (Statement I & II):
Statement I: Primary structure is the linear chain of amino acids.
Statement II: Peptide bonds link successive amino acids.
a) Both I and II are correct
b) Only I is correct
c) Only II is correct
d) Both are incorrect
Explanation: Primary structure refers to the linear amino acid sequence, and peptide bonds link successive amino acids in this chain. Both statements are correct. Correct answer is a) Both I and II are correct.
Chapter: Biomolecules; Topic: Nitrogenous Compounds and Plant Pigments; Subtopic: Purines, Polysaccharides, Alkaloids, Pigments
Keyword Definitions:
Adenine: A purine nitrogenous base found in DNA, RNA, and ATP.
Anthocyanin: Water-soluble pigments responsible for red, purple, and blue colors in plants.
Chitin: A structural polysaccharide present in fungal cell walls and exoskeleton of arthropods.
Codeine: An alkaloid obtained from opium, used as a painkiller and cough suppressant.
Purine: Nitrogenous bases like adenine and guanine forming nucleotides in nucleic acids.
Polysaccharide: Long-chain carbohydrates like chitin, cellulose, and starch.
Alkaloid: Nitrogen-containing naturally occurring organic compounds with physiological effects on humans.
Pigment: Molecules that absorb and reflect light to give color.
Lead Question – 2022 (Ganganagar)
Match List-I with List-II:
List-I List-II
(a) Adenine (i) Pigment
(b) Anthocyanin (ii) Polysaccharide
(c) Chitin (iii) Alkaloid
(d) Codeine (iv) Purine
Choose the correct answer from the options given below:
(a) (b) (c) (d)
1. (iv) (i) (ii) (iii)
2. (iv) (iii) (ii) (i)
3. (iii) (i) (iv) (ii)
4. (i) (iv) (iii) (ii)
Explanation:
Correct answer is option 1. Adenine is a purine base forming nucleotides in DNA and RNA, corresponding to (iv). Anthocyanin is a plant pigment responsible for red, purple, and blue colors, matching (i). Chitin is a structural polysaccharide found in fungal cell walls and arthropod exoskeletons, corresponding to (ii). Codeine is an alkaloid derived from opium with medicinal properties, matching (iii). Understanding the biochemical classification of these compounds is essential for recognizing their roles in nucleic acids, plant physiology, and medicinal chemistry. Each pairing reflects its chemical nature accurately.
1. Single Correct Answer MCQ:
Which of the following is a purine nitrogenous base?
1. Adenine
2. Anthocyanin
3. Chitin
4. Codeine
Explanation: Correct answer is Adenine. Adenine is a purine base present in DNA and RNA. Purines have a double-ring structure and pair with pyrimidines in nucleic acids. Anthocyanin is a pigment, chitin is a polysaccharide, and codeine is an alkaloid, none of which are nitrogenous bases.
2. Single Correct Answer MCQ:
Which compound is responsible for red, purple, and blue colors in plants?
1. Adenine
2. Anthocyanin
3. Chitin
4. Codeine
Explanation: Correct answer is Anthocyanin. Anthocyanins are water-soluble pigments found in vacuoles of plant cells and contribute to colors in flowers, fruits, and leaves. Adenine is a nucleic acid base, chitin is a polysaccharide, and codeine is an alkaloid, which are unrelated to pigmentation.
3. Single Correct Answer MCQ:
Which of the following is a structural polysaccharide?
1. Adenine
2. Anthocyanin
3. Chitin
4. Codeine
Explanation: Correct answer is Chitin. Chitin is a long-chain polymer of N-acetylglucosamine forming structural components in fungi and arthropods. Adenine is a purine base, anthocyanin is a pigment, and codeine is an alkaloid, none of which serve structural functions.
4. Single Correct Answer MCQ:
Which of the following is an alkaloid?
1. Adenine
2. Anthocyanin
3. Chitin
4. Codeine
Explanation: Correct answer is Codeine. Codeine is a nitrogen-containing alkaloid obtained from opium with analgesic and antitussive properties. Adenine is a purine, anthocyanin is a pigment, and chitin is a polysaccharide, which do not have alkaloid properties.
5. Single Correct Answer MCQ:
Purines in nucleic acids include:
1. Adenine and Guanine
2. Cytosine and Thymine
3. Anthocyanin and Chitin
4. Codeine and Anthocyanin
Explanation: Correct answer is Adenine and Guanine. Purines have a double-ring nitrogenous structure and pair with pyrimidines (cytosine, thymine, uracil) in DNA/RNA. Anthocyanin is a pigment, chitin is a polysaccharide, and codeine is an alkaloid, unrelated to nucleic acid structure.
6. Single Correct Answer MCQ:
Which of the following is water-soluble and found in plant vacuoles?
1. Chitin
2. Anthocyanin
3. Adenine
4. Codeine
Explanation: Correct answer is Anthocyanin. Anthocyanins are stored in vacuoles, giving red, purple, or blue color depending on pH. Chitin is a polysaccharide, adenine is a purine base, and codeine is an alkaloid, which are insoluble or function differently.
7. Assertion-Reason MCQ:
Assertion (A): Chitin is a structural polysaccharide.
Reason (R): Chitin is made of repeating units of N-acetylglucosamine forming strong fibers.
1. Both A and R are correct and R is the correct explanation of A
2. Both A and R are correct but R is not the correct explanation of A
3. A is correct, R is false
4. A is false, R is correct
Explanation: Correct answer is option 1. Chitin’s polymer of N-acetylglucosamine forms rigid structures in fungal walls and exoskeletons, making it a structural polysaccharide. The chemical structure directly explains its function in providing strength and protection.
8. Matching Type MCQ:
Match compounds with their type:
A. Adenine – 1. Purine
B. Anthocyanin – 2. Pigment
C. Chitin – 3. Polysaccharide
D. Codeine – 4. Alkaloid
1. A-1, B-2, C-3, D-4
2. A-2, B-1, C-4, D-3
3. A-4, B-3, C-2, D-1
4. A-3, B-4, C-1, D-2
Explanation: Correct answer is option 1. Adenine is a purine base, anthocyanin is a plant pigment, chitin is a polysaccharide, and codeine is an alkaloid. Matching highlights functional and chemical categories of these biomolecules.
9. Fill in the Blanks MCQ:
_______ is an alkaloid obtained from opium used as a painkiller.
1. Adenine
2. Chitin
3. Codeine
Topic: Proteins
Subtopic: Peptide Bond Formation
Amino Acid: Organic molecule containing amino (-NH2) and carboxyl (-COOH) groups; building blocks of proteins.
Carboxyl Group: Functional group (-COOH) present in amino acids involved in peptide bond formation.
Amino Group: Functional group (-NH2) in amino acids that reacts with carboxyl group to form peptide bonds.
Peptide Bond: Covalent bond formed between amino group of one amino acid and carboxyl group of another by dehydration.
Hydrogen Bond: Weak bond formed between a hydrogen atom and electronegative atom like oxygen or nitrogen.
Phosphodiester Bond: Covalent bond linking nucleotides in DNA and RNA between phosphate and sugar moieties.
Glycosidic Bond: Covalent bond connecting sugar molecules in carbohydrates.
Dehydration Reaction: Chemical reaction involving elimination of water to form a new covalent bond.
Protein: Macromolecule composed of amino acids linked via peptide bonds.
Covalent Bond: Strong chemical bond formed by sharing electrons between atoms.
Polypeptide: Chain of amino acids linked by peptide bonds forming protein backbone.
Lead Question - 2022 (Abroad)
Which of the following bond is formed as a result of reaction of carboxyl group of one amino acid with amino group of other amino acid with elimination of water?
Phosphodiester Bond
Hydrogen Bond
Glycosidic Bond
Peptide Bond
Explanation: When the carboxyl group of one amino acid reacts with the amino group of another amino acid, a covalent bond called a peptide bond is formed. This reaction involves elimination of a water molecule, known as a dehydration synthesis. Peptide bonds link amino acids to form proteins. Answer: Peptide Bond. Answer: 4
Q1: Which functional group in amino acids donates a hydrogen for hydrogen bonding in protein secondary structure?
Carboxyl Group
Amino Group
Hydroxyl Group
R Group
Explanation: In protein secondary structures like alpha-helix and beta-sheet, the hydrogen of the amino group (-NH) forms hydrogen bonds with the oxygen of carboxyl group (-C=O) of another amino acid, stabilizing the structure. Answer: Amino Group. Answer: 2
Q2: The bond between sugar and phosphate in DNA is called:
Peptide Bond
Phosphodiester Bond
Glycosidic Bond
Hydrogen Bond
Explanation: In DNA and RNA, nucleotides are linked via phosphodiester bonds between the 3'-OH of one sugar and 5'-phosphate of another. Peptide bonds link amino acids, glycosidic bonds link sugars, and hydrogen bonds form base pairs. Answer: Phosphodiester Bond. Answer: 2
Q3: Which bond is primarily responsible for the tertiary structure of proteins?
Peptide Bond
Disulfide Bond
Hydrogen Bond
Glycosidic Bond
Explanation: Tertiary structure of proteins is stabilized by interactions between R-groups, including disulfide bonds (covalent), hydrogen bonds, ionic bonds, and hydrophobic interactions. Peptide bonds form primary structure, glycosidic bonds are in carbohydrates. Answer: Disulfide Bond. Answer: 2
Q4: Polypeptide chains are formed by repeating:
Glycosidic bonds
Peptide bonds
Phosphodiester bonds
Hydrogen bonds
Explanation: Amino acids are linked by peptide bonds to form polypeptides, which fold into functional proteins. Glycosidic bonds link sugars, phosphodiester bonds link nucleotides, and hydrogen bonds stabilize secondary structures. Answer: Peptide Bonds. Answer: 2
Q5: Which reaction forms a peptide bond?
Hydrolysis
Dehydration synthesis
Oxidation
Reduction
Explanation: Peptide bonds are formed by dehydration synthesis, where a water molecule is eliminated between the amino group of one amino acid and the carboxyl group of another. Hydrolysis breaks the bond. Answer: Dehydration synthesis. Answer: 2
Q6: Which bond type is covalent and links amino acids?
Hydrogen Bond
Disulfide Bond
Peptide Bond
Van der Waals
Explanation: Peptide bonds are covalent bonds linking amino acids to form proteins. Hydrogen bonds are weak interactions, disulfide bonds form between cysteine residues, and Van der Waals interactions are weak non-covalent forces. Answer: Peptide Bond. Answer: 3
Q7: Assertion (A): Proteins are formed by condensation of amino acids.
Reason (R): Peptide bonds are formed with elimination of water between amino acids.
A is correct but R is not correct
A is not correct but R is correct
Both A and R are correct and R explains A
Both A and R are correct but R does not explain A
Explanation: Proteins are polypeptides formed by condensation reactions of amino acids. Peptide bonds are covalent linkages formed by elimination of water between amino and carboxyl groups. Both statements are true, and the reason explains the assertion. Answer: Both A and R are correct and R explains A. Answer: 3
Q8: Match the bond with biomolecule:
1. Peptide Bond A. Proteins
2. Glycosidic Bond B. Carbohydrates
3. Phosphodiester Bond C. Nucleic Acids
1-A, 2-B, 3-C
1-B, 2-A, 3-C
1-C, 2-B, 3-A
1-A, 2-C, 3-B
Explanation: Peptide bonds link amino acids in proteins, glycosidic bonds link monosaccharides in carbohydrates, and phosphodiester bonds link nucleotides in nucleic acids. Correct matching is 1-A, 2-B, 3-C. Answer: 1
Q9: Hydrolysis of a peptide bond requires ______.
Water
ATP
Enzymes only
Heat only
Explanation: Peptide bond hydrolysis breaks the bond between amino acids by adding water, often catalyzed by protease enzymes. ATP or heat alone cannot cleave the bond efficiently. Answer: Water. Answer: 1
Q10: Select correct statements regarding peptide bonds:
Formed between amino and carboxyl groups
Involves elimination of water
Links amino acids to form polypeptides
Is a covalent bond
Explanation: Peptide bonds form covalent linkages between amino and carboxyl groups of amino acids via dehydration reaction, creating polypeptides. All four statements correctly describe peptide bonds. Answer: 1,2,3,4
Subtopic: Biological Molecules and Their Functions
Keyword Definitions:
- Glycogen: Polysaccharide storing glucose in animals.
- Globulin: Plasma protein involved in transport and immunity.
- Steroids: Lipid molecules acting as hormones.
- Thrombin: Enzyme involved in blood clotting.
- Biocatalyst: Substance that speeds up biochemical reactions, usually an enzyme.
- Antibody: Protein that recognizes and neutralizes foreign molecules.
- Hormone: Chemical messenger regulating physiological processes.
- Storage product: Molecules stored in cells for future energy use.
- Polysaccharide: Carbohydrate composed of many monosaccharide units.
- Enzyme: Protein catalyzing biochemical reactions.
- Plasma protein: Proteins in blood with structural, transport, or immune roles.
Lead Question - 2022:
Match List - I with List - II
List - I List - II
(a) Glycogen (i) Hormone
(b) Globulin (ii) Biocatalyst
(c) Steroids (iii) Antibody
(d) Thrombin (iv) Storage product
Choose the correct answer from the options given below:
(1) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
(2) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(3) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
Explanation: The correct answer is (3). Glycogen is a storage product of glucose in animals. Globulin functions as an antibody in immunity. Steroids act as hormones regulating physiological processes. Thrombin is a biocatalyst enzyme responsible for blood clotting. Each molecule is accurately matched with its function.
1. Single Correct Answer:
Which molecule stores glucose in animal cells?
(a) Globulin
(b) Glycogen
(c) Steroids
(d) Thrombin
Explanation: Glycogen is a storage polysaccharide in animals, storing glucose for energy. Globulin transports substances, steroids act as hormones, and thrombin is an enzyme in clotting. Therefore, glycogen is the primary storage product of glucose in animal cells.
2. Single Correct Answer:
Which molecule functions as an antibody?
(a) Glycogen
(b) Steroids
(c) Globulin
(d) Thrombin
Explanation: Globulin includes immunoglobulins, which act as antibodies recognizing and neutralizing pathogens. Glycogen stores energy, steroids act as chemical messengers, and thrombin catalyzes blood clotting. Hence, globulin performs the primary immune defense role as antibodies.
3. Single Correct Answer:
Which molecule acts as a hormone?
(a) Glycogen
(b) Globulin
(c) Steroids
(d) Thrombin
Explanation: Steroids act as hormones regulating physiological processes, including metabolism and reproduction. Glycogen stores glucose, globulin functions as an antibody, and thrombin is a clotting enzyme. Therefore, steroids serve as chemical messengers in the body.
4. Single Correct Answer:
Which is a blood-clotting enzyme?
(a) Glycogen
(b) Thrombin
(c) Globulin
(d) Steroids
Explanation: Thrombin is a biocatalyst enzyme converting fibrinogen into fibrin during blood clotting. Glycogen stores glucose, globulin acts as antibody, and steroids act as hormones. Thrombin’s enzymatic role ensures rapid formation of a clot to prevent excessive bleeding.
5. Single Correct Answer:
Which molecule is a polysaccharide?
(a) Glycogen
(b) Globulin
(c) Thrombin
(d) Steroids
Explanation: Glycogen is a polysaccharide composed of glucose units linked together for energy storage. Globulin and thrombin are proteins, while steroids are lipids. Glycogen’s structure allows rapid mobilization of glucose when the body requires energy.
6. Single Correct Answer:
Which protein functions as a biocatalyst?
(a) Globulin
(b) Thrombin
(c) Glycogen
(d) Steroids
Explanation: Thrombin is an enzyme acting as a biocatalyst in blood coagulation. Globulin functions as an antibody, glycogen stores glucose, and steroids act as hormones. Enzymatic activity of thrombin accelerates the conversion of fibrinogen to fibrin during clot formation.
7. Assertion-Reason MCQ:
Assertion (A): Glycogen is a storage molecule.
Reason (R): It is composed of linked glucose units forming a polysaccharide.
(a) Both A and R are true, R is correct explanation of A
(b) Both A and R are true, R is not correct explanation of A
(c) A is true, R is false
(d) A is false, R is true
Explanation: Option (a) is correct. Glycogen stores glucose for energy in animals. It is a polysaccharide composed of linked glucose units. Both assertion and reason are true, and the reason directly explains why glycogen serves as an energy storage molecule.
8. Matching Type MCQ:
Match molecules with their primary functions:
List - I List - II
(a) Glycogen (i) Storage product
(b) Globulin (ii) Antibody
(c) Steroids (iii) Hormone
(d) Thrombin (iv) Biocatalyst
Options:
(1) a-i, b-ii, c-iii, d-iv
(2) a-ii, b-iv, c-i, d-iii
(3) a-iii, b-i, c-iv, d-ii
(4) a-iv, b-iii, c-ii, d-i
Explanation: Option (1) is correct. Glycogen stores energy, globulin functions as antibodies, steroids act as hormones, and thrombin is a biocatalyst. This mapping accurately aligns molecules with their biological roles.
9. Fill in the Blanks:
________ acts as a chemical messenger regulating physiological processes.
(a) Glycogen
(b) Globulin
(c) Steroids
(d) Thrombin
Explanation: Steroids act as hormones, chemical messengers controlling metabolism, growth, and reproduction. Glycogen stores glucose, globulin acts as antibody, thrombin is a blood-clotting enzyme. Hormonal activity of steroids ensures coordination of complex physiological processes across organs and tissues.
10. Choose the Correct Statements:
Identify correct statements:
1. Glycogen is a storage polysaccharide.
2. Globulin acts as an antibody.
3. Steroids function as biocatalysts.
4. Thrombin is an enzyme involved in clotting.
Options:
(a) 1, 2, 3
(b) 1, 2, 4
(c) 2, 3, 4
(d) 1, 3, 4
Explanation: Option (b) is correct. Glycogen stores glucose, globulin acts as antibody, and thrombin functions as enzyme in clotting. Steroids act as hormones, not enzymes, making statement 3 incorrect. All other statements correctly describe biomolecules and their biological roles.
Chapter: Carbohydrates
Topic: Disaccharides
Subtopic: Formation of Maltose
Keyword Definitions:
Glucose: A simple monosaccharide sugar with formula C6H12O6, primary energy source for cells.
Maltose: A disaccharide formed by two glucose molecules linked via a glycosidic bond.
Dehydration Reaction: A chemical reaction where two molecules combine with loss of water (H2O).
Glycosidic Bond: Covalent bond formed between carbohydrate molecules during disaccharide formation.
Lead Question (2022):
A dehydration reaction links two glucose molecules to produce maltose. If the formula for glucose is C6H12O6, then what is the formula for maltose:
(1) C12H24O12
(2) C12H22O11
(3) C12H24O11
(4) C12H20O10
Explanation: Two glucose molecules (C6H12O6 each) combine via a dehydration reaction, losing one H2O molecule. Thus, C6H12O6 + C6H12O6 – H2O = C12H22O11. Maltose has the formula C12H22O11. Hence, the correct answer is option (2).
1. Maltose is classified as a:
(1) Monosaccharide
(2) Disaccharide
(3) Polysaccharide
(4) Oligosaccharide
Explanation: Maltose is a disaccharide composed of two glucose units joined by an α-1,4-glycosidic bond. Monosaccharides are single sugars, polysaccharides have many units, and oligosaccharides have 3-10 sugar units. Hence, the correct answer is option (2).
2. Which type of bond joins two glucose molecules in maltose?
(1) Peptide bond
(2) Phosphodiester bond
(3) Glycosidic bond
(4) Hydrogen bond
Explanation: Two glucose molecules in maltose are linked by an α-1,4-glycosidic bond formed during dehydration reaction. Peptide bonds join amino acids, phosphodiester bonds join nucleotides, and hydrogen bonds are non-covalent interactions. Hence, the correct answer is option (3).
3. Dehydration synthesis results in:
(1) Formation of water molecule
(2) Breaking of glycosidic bond
(3) Hydrolysis of disaccharide
(4) Release of CO2
Explanation: Dehydration synthesis joins two monosaccharides with the loss of one water molecule (H2O), forming a disaccharide. It is opposite to hydrolysis, which breaks glycosidic bonds. Hence, the correct answer is option (1).
4. Which of the following disaccharides is NOT made of glucose units?
(1) Maltose
(2) Lactose
(3) Cellobiose
(4) Trehalose
Explanation: Lactose is composed of glucose and galactose, whereas maltose, cellobiose, and trehalose are glucose-glucose disaccharides. Hence, the correct answer is option (2).
5. Hydrolysis of maltose produces:
(1) Two fructose molecules
(2) Two glucose molecules
(3) Glucose and galactose
(4) Glucose and fructose
Explanation: Maltose is composed of two glucose units. Hydrolysis of maltose by maltase breaks the glycosidic bond, yielding two glucose molecules. Hence, the correct answer is option (2).
6. In maltose, the glycosidic bond is formed between:
(1) C1 of first glucose and C4 of second glucose
(2) C1 and C6 of same glucose
(3) C2 and C3 of glucose molecules
(4) C4 of first and C4 of second glucose
Explanation: In maltose, the α-1,4-glycosidic bond links C1 of the first glucose to C4 of the second glucose. This bond forms during dehydration synthesis. Hence, the correct answer is option (1).
7. Assertion (A): Maltose is a reducing sugar.
Reason (R): It has a free aldehyde group or hemiacetal functional group.
(1) Both A and R are true and R is correct explanation
(2) Both A and R are true but R is not correct explanation
(3) A is true, R is false
(4) A is false, R is true
Explanation: Maltose is a reducing sugar because the anomeric carbon of the second glucose is free and can form an aldehyde group in equilibrium. This allows it to reduce mild oxidizing agents. Hence, the correct answer is option (1).
8. Matching Type: Match disaccharides with monosaccharide components:
A. Maltose – 1. Glucose + Galactose
B. Lactose – 2. Glucose + Glucose
C. Sucrose – 3. Glucose + Fructose
(1) A–2, B–1, C–3
(2) A–1, B–2, C–3
(3) A–3, B–1, C–2
(4) A–2, B–3, C–1
Explanation: Maltose is glucose + glucose, lactose is glucose + galactose, and sucrose is glucose + fructose. Hence, the correct answer is option (1).
9. Fill in the blanks:
The disaccharide formed from two glucose molecules via a dehydration reaction is __________.
(1) Sucrose
(2) Lactose
(3) Maltose
(4) Trehalose
Explanation: Two glucose molecules combine via dehydration synthesis to form malt
Topic: DNA Structure
Subtopic: DNA Length and Base Pair Relationship
Keyword Definitions:
DNA: Deoxyribonucleic acid, a molecule that carries genetic information in living organisms.
Base Pair (bp): A pair of nitrogenous bases connected by hydrogen bonds in a DNA molecule, such as A–T and G–C.
Nucleotide: Basic unit of DNA containing a sugar, phosphate group, and nitrogen base.
Double Helix: The coiled structure of DNA with two complementary strands wound around each other.
Genomic DNA: Total genetic material contained within an organism’s chromosomes.
Lead Question – 2022
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs:
(1) 6.6×109 bp
(2) 3.3×106 bp
(3) 6.6×106 bp
(4) 3.3×109 bp
Explanation: The distance between two consecutive base pairs is 0.34 nanometres. Therefore, 1.1 m DNA equals 1.1×109 nm. Dividing by 0.34 nm per bp gives approximately 3.3×109 base pairs. Hence, the correct answer is (4).
1. In a DNA double helix, the two strands are:
(1) Parallel and identical
(2) Antiparallel and complementary
(3) Antiparallel and identical
(4) Parallel and complementary
Explanation: In DNA, one strand runs 5′ to 3′ and the other 3′ to 5′. Their base sequences are complementary (A–T, G–C). Thus, the two strands are antiparallel and complementary. Hence, the correct answer is (2).
2. The distance between two consecutive base pairs in DNA is:
(1) 0.34 nm
(2) 3.4 nm
(3) 34 nm
(4) 0.034 nm
Explanation: Each base pair in the DNA double helix is separated by a distance of 0.34 nanometres. Thus, ten base pairs together form one complete turn measuring 3.4 nm. Hence, the correct answer is (1).
3. The number of base pairs in one complete turn of B-DNA is:
(1) 5
(2) 10
(3) 15
(4) 20
Explanation: In B-form DNA, there are 10 base pairs per complete turn of the helix, and each turn measures 3.4 nm. Hence, the correct answer is (2).
4. The length of a human diploid cell DNA is approximately:
(1) 1.1 m
(2) 2.2 m
(3) 3.3 m
(4) 6.6 m
Explanation: A haploid human genome (one set of chromosomes) contains about 3.3×109 base pairs, corresponding to 1.1 m DNA. Hence, a diploid cell with two sets has approximately 2.2 m DNA. Correct answer is (2).
5. If 200 base pairs of DNA make one turn, what is its total length?
(1) 68 nm
(2) 6.8 nm
(3) 0.68 μm
(4) 0.068 μm
Explanation: Each base pair contributes 0.34 nm to the DNA length. Thus, 200 × 0.34 = 68 nm. Therefore, total length = 68 nm. Hence, the correct answer is (1).
6. Which of the following statements about DNA is correct?
(1) It contains uracil instead of thymine
(2) Its sugar component is ribose
(3) Its backbone is made of phosphate and deoxyribose sugar
(4) It is single-stranded in all organisms
Explanation: DNA is made up of deoxyribose sugar and phosphate forming its backbone, while bases (A, T, G, C) are attached. Uracil is found in RNA, not DNA. Hence, the correct answer is (3).
7. Assertion (A): DNA strands are antiparallel.
Reason (R): The hydrogen bonds form only when one strand runs 5′→3′ and the other 3′→5′.
(1) Both A and R are true, and R explains A
(2) Both A and R are true, but R does not explain A
(3) A is true, R is false
(4) A is false, R is true
Explanation: The hydrogen bonding between bases is possible only when strands are oriented oppositely (antiparallel). Thus, both A and R are true, and R correctly explains A. Hence, the correct answer is (1).
8. Match the following:
A. Hydrogen bond — (i) Adenine–Thymine
B. Phosphodiester bond — (ii) Between sugars and phosphates
C. Complementary base pairing — (iii) DNA stability
D. Double helix — (iv) Watson and Crick
Options:
(1) A-(i), B-(ii), C-(iii), D-(iv)
(2) A-(ii), B-(i), C-(iv), D-(iii)
(3) A-(iv), B-(iii), C-(ii), D-(i)
(4) A-(iii), B-(i), C-(iv), D-(ii)
Explanation: Hydrogen bonds link complementary bases (A–T, G–C), phosphodiester bonds link sugar-phosphate backbone, and the double helix structure was proposed by Watson and Crick. Hence, the correct answer is (1).
9. Fill in the blank:
Each complete turn of B-DNA measures _______ nm.
(1) 0.34
(2) 3.4
(3) 34
(4) 0.034
Explanation: In the B-form of DNA, there are 10 base pairs per turn, and the distance between adjacent base pairs is 0.34 nm. Hence, one complete turn measures 3.4 nm. Correct answer is (2).
10. Choose the correct statements:
(1) A–T pairs have three hydrogen bonds.
(2) G–C pairs have two hydrogen bonds.
(3) Base pairing follows Chargaff’s rule.
(4) A–T pairs have two hydrogen bonds.
Explanation: According to Chargaff’s rule, A pairs with T via two hydrogen bonds, and G pairs with C via three hydrogen bonds. This complementary pairing ensures structural stability of DNA. Hence, statements (3) and (4) are correct.
Topic: Lipids and Fatty Acids
Subtopic: Structure, Properties, and Biological Importance of Lipids
Keyword Definitions:
Lipids: Organic compounds that are insoluble in water but soluble in organic solvents.
Fatty acids: Long hydrocarbon chains with a carboxylic acid group.
Saturated fatty acids: Fatty acids without double bonds between carbon atoms.
Unsaturated fatty acids: Fatty acids with one or more double bonds (C=C).
Lecithin: A phospholipid found in cell membranes, not a glycolipid.
Monoglycerides: Formed by esterification of one fatty acid molecule with glycerol.
Lead Question (2022):
Read the following statements on lipids and find out correct set of statements.
(a) Lecithin found in the plasma membrane is a glycolipid
(b) Saturated fatty acids possess one or more C=C bonds
(c) Gingely oil has lower melting point, hence remains as oil in winter
(d) Lipids are generally insoluble in water but soluble in some organic solvents
(e) When fatty acid is esterified with glycerol, monoglycerides are formed
Choose the correct answer from the options given below:
(1) (a), (d), and (e) only
(2) (c), (d), and (e) only
(3) (a), (b), and (d) only
(4) (a), (b), and (c) only
Explanation: Lecithin is a phospholipid, not a glycolipid. Saturated fatty acids lack double bonds, while gingely oil (unsaturated) remains liquid due to low melting point. Lipids are insoluble in water but dissolve in organic solvents. Fatty acids esterify with glycerol to form monoglycerides. Hence, correct answer is (2) (c), (d), and (e) only.
1. Guessed Question:
Which of the following is a derived lipid?
(1) Waxes
(2) Steroids
(3) Triglycerides
(4) Phospholipids
Explanation: Derived lipids are obtained by hydrolysis of simple and compound lipids. Steroids, including cholesterol and hormones, belong to this group. They play crucial roles in membrane structure and signaling. Hence, the correct answer is (2) Steroids.
2. Guessed Question:
Which statement about unsaturated fatty acids is correct?
(1) They have higher melting points
(2) They are solid at room temperature
(3) They contain double bonds
(4) They are completely hydrogenated
Explanation: Unsaturated fatty acids contain one or more double bonds in their hydrocarbon chain, resulting in kinks that lower melting points. Thus, they remain liquid at room temperature, as seen in vegetable oils. Hence, the correct answer is (3) They contain double bonds.
3. Guessed Question:
Which of the following is NOT a simple lipid?
(1) Fats
(2) Oils
(3) Waxes
(4) Phospholipids
Explanation: Simple lipids include fats, oils, and waxes formed by fatty acids and alcohol. Phospholipids are compound lipids, containing phosphorus, nitrogen, and other groups. Hence, the correct answer is (4) Phospholipids.
4. Guessed Question:
Which of the following lipid functions is correct?
(1) Structural component of cell wall
(2) Long-term energy storage
(3) Genetic information storage
(4) Enzyme catalysis
Explanation: Lipids serve as long-term energy reserves, providing more energy per gram than carbohydrates. They also form cellular membranes and act as signaling molecules. Hence, the correct answer is (2) Long-term energy storage.
5. Guessed Question:
Which of the following lipids is amphipathic?
(1) Cholesterol
(2) Phospholipids
(3) Triglycerides
(4) Sterols
Explanation: Phospholipids possess hydrophilic (polar head) and hydrophobic (fatty acid tails) regions, making them amphipathic. This property allows them to form lipid bilayers in membranes. Hence, the correct answer is (2) Phospholipids.
6. Guessed Question:
Which statement about triglycerides is TRUE?
(1) They are polymers of glucose
(2) They consist of glycerol and three fatty acids
(3) They contain phosphates
(4) They are hydrophilic
Explanation: Triglycerides are formed by the esterification of glycerol with three fatty acid molecules. They are hydrophobic and serve as major energy storage molecules in animals. Hence, the correct answer is (2) They consist of glycerol and three fatty acids.
7. Assertion-Reason Question:
Assertion (A): Saturated fats are solid at room temperature.
Reason (R): They have no double bonds between carbon atoms.
(1) Both A and R are true and R explains A.
(2) Both A and R are true but R does not explain A.
(3) A is true but R is false.
(4) A is false but R is true.
Explanation: Saturated fats have tightly packed molecules due to absence of double bonds, resulting in higher melting points and solid form at room temperature. Hence, both A and R are true and R correctly explains A. Correct answer is (1).
8. Matching Type Question:
Match the following lipids with their functions:
A. Cholesterol — (i) Energy storage
B. Phospholipids — (ii) Membrane structure
C. Triglycerides — (iii) Hormone precursor
Options:
(1) A–iii, B–ii, C–i
(2) A–ii, B–iii, C–i
(3) A–i, B–ii, C–iii
(4) A–iii, B–i, C–ii
Explanation: Cholesterol acts as a hormone precursor, phospholipids contribute to membrane structure, and triglycerides store energy. Hence, the correct answer is (1) A–iii, B–ii, C–i.
9. Fill in the Blanks:
Lipids are composed mainly of ______ and ______.
(1) Carbon and hydrogen
(2) Oxygen and nitrogen
(3) Nitrogen and sulfur
(4) Carbon and phosphorus
Explanation: Lipids are primarily composed of carbon and hydrogen atoms with smaller amounts of oxygen. This composition makes them non-polar and insoluble in water. Hence, the correct answer is (1) Carbon and hydrogen.
10. Choose the Correct Statements:
(a) Lipids are hydrophilic compounds.
(b) Steroids are examples of derived lipids.
(c) Fats yield more energy than carbohydrates.
(d) Lecithin is a glycolipid.
(1) (a) and (d) only
(2) (b) and (c) only
(3) (a), (b), and (c) only
(4) (b), (c), and (d) only
Explanation: Lipids are hydrophobic, not hydrophilic. Steroids are derived lipids, and fats provide more energy than carbohydrates. Lecithin is a phospholipid, not a glycolipid. Hence, the correct answer is (2) (b) and (c) only.
Topic: Lipids
Subtopic: Types and Properties of Lipids
Keyword Definitions:
Lipids: Organic molecules that are hydrophobic, include fats, oils, phospholipids, and steroids, serving as energy storage and structural components.
Unsaturated fatty acids: Fatty acids containing one or more double bonds between carbon atoms.
Phospholipids: Lipids containing a phosphate group, forming cell membrane components like lecithin.
Glycerol: Trihydroxy alcohol forming the backbone of triglycerides and phospholipids.
Palmitic acid: A saturated fatty acid with 16 carbon atoms including carboxyl carbon.
Arachidonic acid: A polyunsaturated fatty acid with 20 carbon atoms.
Lead Question - 2021
Following are the statements with reference to 'lipids'.
(a) Lipids having only single bonds are called unsaturated fatty acids.
(b) Lecithin is a phospholipid
(c) Trihydroxy propane is glycerol.
(d) Palmitic acid has 20 carbon atoms including carboxyl carbon
(e) Arachidonic acid has 16 carbon atoms
Choose the correct answer from the options given below
(1) (c) and (d) only
(2) (b) and (c) only
(3) (b) and (e) only
(4) (a) and (b) only
Explanation: Statement (a) is incorrect; lipids with only single bonds are saturated fatty acids. Lecithin is a phospholipid (b), and trihydroxy propane is glycerol (c). Palmitic acid has 16 carbons, not 20, and arachidonic acid has 20 carbons, not 16. Correct answer is option (2) (b) and (c) only.
1. Which fatty acid has one double bond?
(1) Palmitic acid
(2) Oleic acid
(3) Stearic acid
(4) Lauric acid
Explanation: Oleic acid is a monounsaturated fatty acid with one double bond, while palmitic, stearic, and lauric acids are saturated. Monounsaturated fatty acids influence membrane fluidity and health benefits. Correct answer is option (2) Oleic acid.
2. Which component forms the hydrophilic head of phospholipids?
(1) Fatty acid
(2) Glycerol
(3) Phosphate group
(4) Triglyceride
Explanation: The phosphate group forms the polar hydrophilic head of phospholipids, while fatty acids form hydrophobic tails. Glycerol acts as the backbone connecting the head and tails. Correct answer is option (3) Phosphate group.
3. Which lipid is a primary energy storage molecule?
(1) Phospholipid
(2) Triglyceride
(3) Steroid
(4) Lecithin
Explanation: Triglycerides store energy efficiently in adipose tissue, composed of glycerol and three fatty acids. Phospholipids are structural, steroids regulate metabolism, and lecithin is a phospholipid. Correct answer is option (2) Triglyceride.
4. Which lipid forms the bilayer of cell membranes?
(1) Steroids
(2) Triglycerides
(3) Phospholipids
(4) Wax
Explanation: Phospholipids arrange into bilayers forming the fundamental structure of cell membranes, with hydrophilic heads outside and hydrophobic tails inside. Steroids modulate fluidity, triglycerides store energy, waxes provide protection. Correct answer is option (3) Phospholipids.
5. Which lipid contains four fused rings?
(1) Triglyceride
(2) Phospholipid
(3) Steroid
(4) Glycolipid
Explanation: Steroids have a structure of four fused hydrocarbon rings, examples include cholesterol and hormones. Triglycerides are glycerol + fatty acids, phospholipids form membranes, glycolipids are membrane lipids with sugar. Correct answer is option (3) Steroid.
6. Which fatty acid is polyunsaturated?
(1) Stearic acid
(2) Palmitic acid
(3) Arachidonic acid
(4) Lauric acid
Explanation: Arachidonic acid is a polyunsaturated fatty acid with four double bonds, important in eicosanoid synthesis. Stearic, palmitic, and lauric acids are saturated fatty acids. Correct answer is option (3) Arachidonic acid.
7. Assertion-Reason Question:
Assertion (A): Glycerol is the backbone of triglycerides.
Reason (R): Triglycerides consist of three fatty acids esterified to a trihydroxy alcohol.
(1) Both A and R are true, R is correct explanation of A
(2) Both A and R are true, R is not correct explanation of A
(3) A true, R false
(4) A false, R true
Explanation: Glycerol, a trihydroxy alcohol, serves as the backbone for triglycerides, linking three fatty acids via ester bonds. This makes the reason correct explanation of the assertion. Correct answer is option (1).
8. Matching Type Question:
Match lipid types with examples:
(a) Saturated fatty acid - 1. Oleic acid
(b) Polyunsaturated fatty acid - 2. Arachidonic acid
(c) Phospholipid - 3. Lecithin
(d) Steroid - 4. Cholesterol
(1) a-1, b-2, c-3, d-4
(2) a-2, b-1, c-4, d-3
(3) a-4, b-3, c-2, d-1
(4) a-3, b-4, c-1, d-2
Explanation: Saturated fatty acid example: palmitic acid (a-1), polyunsaturated: arachidonic acid (b-2), phospholipid: lecithin (c-3), steroid: cholesterol (d-4). Correct answer is option (1).
9. Fill in the Blanks:
______ is a trihydroxy alcohol forming the backbone of triglycerides.
(1) Glycerol
(2) Palmitic acid
(3) Oleic acid
(4) Lecithin
Explanation: Glycerol, a trihydroxy alcohol, forms the backbone of triglycerides by esterifying three fatty acids. Palmitic and oleic acids are fatty acids, lecithin is a phospholipid. Correct answer is option (1) Glycerol.
10. Choose the correct statements:
(a) Saturated fatty acids have single bonds
(b) Unsaturated fatty acids have one or more double bonds
(c) Phospholipids form cell membranes
(d) Arachidonic acid has 16 carbon atoms
(1) a, b, c
(2) a, c, d
(3) b, c, d
(4) a, b, d
Explanation: Saturated fatty acids have single bonds (a), unsaturated fatty acids have one or more double bonds (b), and phospholipids form cell membranes (c). Arachidonic acid has 20 carbons, not 16 (d incorrect). Correct answer is option (1) a, b, c.
Subtopic: Base Pairing
DNA: Deoxyribonucleic acid, a molecule carrying genetic instructions.
Adenine (A): A purine base in DNA that pairs with thymine.
Thymine (T): A pyrimidine base in DNA that pairs with adenine.
Guanine (G): A purine base in DNA that pairs with cytosine.
Cytosine (C): A pyrimidine base in DNA that pairs with guanine.
Base pairing: Complementary pairing of DNA bases: A-T and G-C.
Purine: Double-ring nitrogenous base (A and G) in DNA.
Pyrimidine: Single-ring nitrogenous base (T and C) in DNA.
Chargaff's rule: In DNA, %A=%T and %G=%C.
Genetic code: Sequence of nucleotides that determines traits.
Double helix: DNA structure with two complementary strands.
Lead Question - 2021
If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?
(1) T: 20 ; G: 20 ; C: 30
(2) T: 30 ; G: 20 ; C: 20
(3) T: 20 ; G: 25 ; C: 25
(4) T: 20 ; G: 30 ; C: 20
Explanation: According to Chargaff's rule, %A = %T and %G = %C. If Adenine is 30%, Thymine is also 30%. Total percentage of A+T = 60%, so remaining 40% is shared equally by G and C, giving G=20% and C=20%. Answer: T:30; G:20; C:20.
1. Single Correct Answer MCQ: Which base pairs with Guanine in DNA?
Options:
A. Adenine
B. Thymine
C. Cytosine
D. Uracil
Explanation: In DNA, Guanine pairs with Cytosine via three hydrogen bonds. Adenine pairs with Thymine. Uracil is found in RNA, not DNA. Answer: Cytosine.
2. Single Correct Answer MCQ: If Cytosine is 15%, what is the percentage of Guanine?
Options:
A. 15
B. 30
C. 35
D. 20
Explanation: Chargaff's rule states %C=%G. If Cytosine is 15%, Guanine is also 15%. Answer: 15.
3. Single Correct Answer MCQ: Total percentage of purines in DNA if A=30% and G=20% is:
Options:
A. 50%
B. 40%
C. 60%
D. 30%
Explanation: Purines are Adenine and Guanine. A=30%, G=20%, total purines = 30+20 = 50%. Answer: 50%.
4. Single Correct Answer MCQ: What is the percentage of Pyrimidines if A=30% and G=20%?
Options:
A. 30%
B. 50%
C. 40%
D. 20%
Explanation: Pyrimidines are Thymine and Cytosine. %T=%A=30%, %C=%G=20%, total = 30+20 = 50%. Answer: 50%.
5. Single Correct Answer MCQ: Which rule determines base pairing in DNA?
Options:
A. Mendel’s law
B. Chargaff’s rule
C. Hardy-Weinberg principle
D. Law of segregation
Explanation: Chargaff’s rule specifies complementary base pairing: %A=%T, %G=%C in DNA. Other laws are unrelated to base pairing. Answer: Chargaff’s rule.
6. Single Correct Answer MCQ: Which base is not a pyrimidine?
Options:
A. Cytosine
B. Thymine
C. Adenine
D. Uracil
Explanation: Pyrimidines include Cytosine, Thymine, and Uracil. Adenine is a purine. Answer: Adenine.
7. Assertion-Reason MCQ:
Assertion (A): DNA has equal amounts of A and T.
Reason (R): Purine always pairs with a pyrimidine.
Options:
A. Both A and R true, R correct explanation
B. Both A and R true, R not correct explanation
C. A true, R false
D. A false, R true
Explanation: In DNA, A=T and G=C. Purines (A,G) pair with pyrimidines (T,C). Both assertion and reason are true, and R explains A correctly. Answer: Both A and R true, R correct explanation.
8. Matching Type MCQ:
List I: 1. Purines 2. Pyrimidines 3. Base pairing 4. Double helix
List II: A. A-T, G-C B. A,G C. C,T D. Two strands coiled
Options:
A. 1-B, 2-C, 3-A, 4-D
B. 1-C, 2-B, 3-A, 4-D
C. 1-B, 2-C, 3-D, 4-A
D. 1-A, 2-C, 3-B, 4-D
Explanation: Purines are A and G, Pyrimidines are C and T, Base pairing: A-T, G-C, Double helix: two coiled strands. Answer: 1-B, 2-C, 3-A, 4-D.
9. Fill in the Blank MCQ: The percentage of Thymine equals ______.
Options:
A. Adenine
B. Guanine
C. Cytosine
D. Total purines
Subtopic: Bonds in Biomolecules
Keyword Definitions:
Peptide bond: Covalent bond formed between two amino acids in proteins.
Glycosidic bond: Covalent bond that joins monosaccharides to form polysaccharides.
Phosphodiester bond: Linkage connecting nucleotides in nucleic acids.
Unsaturated fatty acid: Fatty acid containing one or more C=C double bonds.
Macromolecules: Large molecules essential for life including proteins, nucleic acids, carbohydrates, and lipids.
Lead Question - 2021
Match List - I with List - II.
List - I List - II
(a) Protein (i) C=C double bonds
(b) Unsaturated fatty acid (ii) Phosphodiester bonds
(c) Nucleic acid (iii) Glycosidic bonds
(d) Polysaccharide (iv) Peptide bonds
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (i) (iv) (iii) (ii)
(2) (ii) (i) (iv) (iii)
(3) (iv) (iii) (i) (ii)
(4) (iv) (i) (ii) (iii)
Explanation: Proteins are linked by peptide bonds, unsaturated fatty acids by C=C double bonds, nucleic acids by phosphodiester bonds, and polysaccharides by glycosidic bonds. Thus, the correct matching is (a)–(iv), (b)–(i), (c)–(ii), (d)–(iii). The correct answer is (4).
1. Which bond is responsible for linking amino acids in a polypeptide chain?
(1) Hydrogen bond
(2) Glycosidic bond
(3) Peptide bond
(4) Phosphodiester bond
Explanation: In proteins, amino acids are joined by peptide bonds formed between the amino group of one amino acid and the carboxyl group of another. This bond is covalent and provides primary structure stability. Thus, the correct answer is (3) Peptide bond.
2. Which bond connects monosaccharides to form polysaccharides like starch?
(1) Hydrogen bond
(2) Peptide bond
(3) Glycosidic bond
(4) Phosphodiester bond
Explanation: Polysaccharides such as starch, glycogen, and cellulose are formed by glycosidic bonds between monosaccharide units. These bonds are covalent and help in energy storage and structural roles. Thus, the correct answer is (3) Glycosidic bond.
3. Assertion (A): Phosphodiester bonds link nucleotides in DNA.
Reason (R): They connect phosphate of one nucleotide to sugar of another.
(1) Both A and R are true, and R is the correct explanation of A.
(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, R is false.
(4) A is false, R is true.
Explanation: Phosphodiester bonds form between the phosphate group of one nucleotide and the hydroxyl group of the sugar in another nucleotide, maintaining DNA and RNA backbone. Both assertion and reason are true and R explains A. Thus, the correct answer is (1).
4. Match the following molecules with their bonds:
A. Proteins
B. Polysaccharides
C. Nucleic acids
D. Lipids
(1) A–Peptide, B–Glycosidic, C–Phosphodiester, D–C=C double bond
(2) A–Glycosidic, B–Peptide, C–Phosphodiester, D–Hydrogen
(3) A–Phosphodiester, B–Peptide, C–Glycosidic, D–Hydrogen
(4) A–Hydrogen, B–Peptide, C–C=C double bond, D–Glycosidic
Explanation: Proteins are held by peptide bonds, polysaccharides by glycosidic bonds, nucleic acids by phosphodiester bonds, and lipids by C=C double bonds if unsaturated. Thus, the correct answer is (1).
5. Fill in the blank:
__________ bonds link nucleotides in nucleic acids.
(1) Glycosidic
(2) Peptide
(3) Phosphodiester
(4) Hydrogen
Explanation: In DNA and RNA, nucleotides are joined by phosphodiester bonds, which form the sugar-phosphate backbone essential for genetic material stability. Thus, the correct answer is (3) Phosphodiester.
6. Which type of bond is present in unsaturated fatty acids?
(1) Peptide bond
(2) Glycosidic bond
(3) Hydrogen bond
(4) C=C double bond
Explanation: Unsaturated fatty acids contain one or more carbon–carbon double bonds (C=C) in their hydrocarbon chains. These bonds influence fluidity and properties of cell membranes. Thus, the correct answer is (4) C=C double bond.
7. Which of the following correctly pairs the biomolecule with its primary bond?
(1) Protein – Glycosidic
(2) Nucleic acid – Phosphodiester
(3) Polysaccharide – Peptide
(4) Lipid – Peptide
Explanation: Nucleic acids such as DNA and RNA have nucleotides linked by phosphodiester bonds in their backbone. Proteins, polysaccharides, and lipids have different bonding types. Thus, the correct answer is (2) Nucleic acid – Phosphodiester.
8. Choose the correct statements about biomolecular bonds:
(1) Proteins are formed by peptide bonds.
(2) Polysaccharides are formed by glycosidic bonds.
(3) Nucleic acids are formed by phosphodiester bonds.
(4) Lipids contain C=C double bonds if unsaturated.
(1) 1 and 2 only
(2) 2 and 4 only
(3) 1, 2 and 3 only
(4) 1, 2, 3 and 4
Explanation: All listed statements are correct. Proteins – peptide bonds, polysaccharides – glycosidic bonds, nucleic acids – phosphodiester bonds, lipids – C=C double bonds if unsaturated. Thus, the correct answer is (4) 1, 2, 3 and 4.
9. In proteins, peptide bonds are formed between:
(1) Carboxyl group of one amino acid and amino group of another
(2) Hydroxyl groups of two amino acids
(3) Phosphate and hydroxyl groups
(4) Sugar and nitrogenous base
Explanation: Peptide bonds form between the carboxyl group of one amino acid and the amino group of another, releasing a molecule of water in a condensation reaction. Thus, the correct answer is (1).
10. Which bond is most stable and resistant to degradation?
(1) Glycosidic bond
(2) Peptide bond
(3) Phosphodiester bond
(4) C=C double bond
Explanation: Among the listed, phosphodiester bonds in DNA are highly stable, enabling long-term storage of genetic information. This stability is essential for heredity. Thus, the correct answer is (3) Phosphodiester bond.
Topic: DNA Structure and Analysis
Subtopic: Gel Electrophoresis and DNA Visualization
Keyword Definitions:
DNA: Deoxyribonucleic acid, the molecule carrying genetic instructions.
Ethidium Bromide: A fluorescent dye that intercalates between DNA bases for visualization under UV light.
Gel Electrophoresis: A technique to separate DNA fragments by size using an electric field in agarose gel.
UV Radiation: Ultraviolet light used to visualize DNA bound to ethidium bromide.
DNA Bands: Distinct visual representations of DNA fragments on a gel.
Intercalation: The insertion of molecules like ethidium bromide between DNA base pairs.
Agarose Gel: A porous medium for electrophoretic separation of nucleic acids.
Fluorescence: Emission of visible light by a substance after absorbing light.
Lead Question - 2021
DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as:
(1) Bright orange bands
(2) Dark red bands
(3) Bright blue bands
(4) Yellow bands
Explanation: Correct answer is (1) Bright orange bands. Ethidium bromide intercalates into DNA and fluoresces bright orange under UV light, allowing visualization of DNA fragments. Other colors do not accurately represent ethidium bromide-DNA fluorescence, which is a standard technique in molecular biology labs for gel electrophoresis.
Guessed Questions:
1) Single Correct Answer: Which dye intercalates between DNA bases for visualization?
(1) Methylene Blue
(2) Ethidium Bromide
(3) Coomassie Blue
(4) Crystal Violet
Explanation: Correct answer is (2) Ethidium Bromide. It binds between DNA base pairs and fluoresces under UV light, enabling visualization of DNA in agarose gels.
2) Single Correct Answer: UV light is used in gel electrophoresis to:
(1) Separate DNA fragments
(2) Visualize stained DNA
(3) Denature RNA
(4) Break proteins
Explanation: Correct answer is (2) Visualize stained DNA. UV light excites ethidium bromide bound to DNA, causing fluorescence, which allows observation of DNA bands.
3) Single Correct Answer: In agarose gel electrophoresis, DNA moves because of:
(1) Diffusion
(2) Electric field
(3) Gravity
(4) Centrifugal force
Explanation: Correct answer is (2) Electric field. DNA is negatively charged due to its phosphate backbone and migrates towards the positive electrode through the porous agarose matrix.
4) Single Correct Answer: DNA bands on gel represent:
(1) Protein fragments
(2) DNA fragment sizes
(3) RNA sequences
(4) Gel density
Explanation: Correct answer is (2) DNA fragment sizes. Each band corresponds to DNA fragments of a specific length, separated during electrophoresis.
5) Single Correct Answer: Fluorescence of DNA stained with ethidium bromide is due to:
(1) UV absorption and emission
(2) Heat emission
(3) Chemical reaction
(4) Light scattering
Explanation: Correct answer is (1) UV absorption and emission. Ethidium bromide absorbs UV light and emits visible orange fluorescence, enabling DNA visualization in gels.
6) Assertion-Reason:
A: DNA fluoresces bright orange under UV when stained with ethidium bromide.
R: Ethidium bromide intercalates between DNA bases and emits visible light.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Correct answer is (1). The fluorescence occurs because ethidium bromide intercalates DNA and emits orange light upon UV excitation, directly explaining the observed bright orange bands.
7) Matching Type: Match the dye with its characteristic:
List-I List-II
(a) Ethidium Bromide (i) DNA intercalating, fluoresces orange
(b) Methylene Blue (ii) Stains DNA blue, non-fluorescent
(c) Coomassie Blue (iii) Stains proteins
(d) Crystal Violet (iv) Stains bacterial cells
Explanation: Correct answer: (a) i, (b) ii, (c) iii, (d) iv. Each dye has a specific binding property and visualization method, commonly used in molecular biology.
8) Single Correct Answer: Ethidium bromide is avoided in some labs due to:
(1) High cost
(2) Toxicity and mutagenicity
(3) Poor staining
(4) Low fluorescence
Explanation: Correct answer is (2) Toxicity and mutagenicity. Ethidium bromide can intercalate into genomic DNA and is a potential mutagen, so safe handling is required.
9) Fill in the blank: DNA stained with ethidium bromide fluoresces ______ under UV light.
(1) Blue
(2) Bright orange
(3) Green
(4) Red
Explanation: Correct answer is (2) Bright orange. Ethidium bromide intercalates DNA and emits bright orange fluorescence under UV light, enabling band detection in gels.
10) Choose the correct statements:
(a) Ethidium bromide binds between DNA bases.
(b) DNA appears bright orange under UV after staining.
(c) Ethidium bromide stains proteins primarily.
(d) DNA migration is towards positive electrode.
Options:
(1) a, b, d
(2) a, c, d
(3) b, c, d
(4) a, b, c
Explanation: Correct answer is (1). Statements a, b, and d are correct. Statement c is incorrect; ethidium bromide primarily stains DNA, not proteins.
Subtopic: Primary vs Secondary Metabolites
Keyword Definitions:
Primary metabolites: Compounds like amino acids, glucose essential for growth and metabolism.
Secondary metabolites: Non-essential compounds like alkaloids, terpenoids, phenolics with ecological functions.
Alkaloids: Nitrogen-containing compounds such as morphine and codeine with medicinal properties.
Terpenoids: Compounds like rubber and essential oils derived from isoprene units.
Phenolics: Compounds like curcumin with antioxidant properties.
Medicinal metabolites: Compounds such as vinblastine with pharmaceutical use.
Gums: Polysaccharides secreted by plants, used in industrial and pharmaceutical processes.
Lead Question - 2021
Which of the following are not secondary metabolites in plants?
(1) Amino acids, glucose
(2) Vinblastin, curcumin
(3) Rubber, gums
(4) Morphine, codeine
Explanation: The correct answer is (1) Amino acids, glucose. These are primary metabolites essential for plant survival and cellular metabolism. Secondary metabolites such as morphine, vinblastine, and curcumin are not directly required for growth but play protective, ecological, or commercial roles. Thus, amino acids and glucose belong to primary, not secondary, metabolites.
Guessed Questions:
1) Which of the following is a secondary metabolite?
(1) Glucose
(2) Amino acid
(3) Morphine
(4) ATP
Explanation: The correct answer is (3) Morphine. It is an alkaloid and secondary metabolite used medicinally. Primary metabolites like glucose, amino acids, and ATP are essential for growth, whereas secondary metabolites like morphine function in defense and ecological interactions without being vital for plant survival.
2) Curcumin, found in turmeric, is chemically classified as:
(1) Alkaloid
(2) Terpenoid
(3) Phenolic compound
(4) Polysaccharide
Explanation: The correct answer is (3) Phenolic compound. Curcumin is a secondary metabolite with antioxidant and medicinal properties. It is not an alkaloid or terpenoid. Instead, it belongs to phenolics derived from plant biosynthetic pathways and is widely used in traditional medicine and food industries.
3) Rubber produced by plants belongs to:
(1) Alkaloids
(2) Terpenoids
(3) Phenolics
(4) Glycosides
Explanation: The correct answer is (2) Terpenoids. Rubber is a polymer of isoprene units and classified under terpenoids. These secondary metabolites protect plants and have vast industrial applications. Unlike alkaloids and phenolics, terpenoids like rubber contribute to defense and elasticity in plants.
4) Which of the following is an example of an alkaloid?
(1) Starch
(2) Morphine
(3) Cellulose
(4) Pectin
Explanation: The correct answer is (2) Morphine. Alkaloids are nitrogen-containing secondary metabolites. Morphine acts as a potent analgesic derived from the opium poppy. Compounds like starch, cellulose, and pectin are primary metabolites involved in structural and energy storage functions.
5) Gums secreted by plants are classified as:
(1) Proteins
(2) Secondary metabolites
(3) Lipids
(4) Enzymes
Explanation: The correct answer is (2) Secondary metabolites. Plant gums are polysaccharides secreted in response to injury or stress. Though not vital for basic metabolism, they provide ecological and industrial benefits. They differ from proteins, lipids, or enzymes in structure and function.
6) Which of the following is a primary metabolite?
(1) Glucose
(2) Curcumin
(3) Morphine
(4) Rubber
Explanation: The correct answer is (1) Glucose. It is a primary metabolite directly involved in cellular respiration and energy production. Secondary metabolites like curcumin, morphine, and rubber have ecological, protective, or commercial roles rather than being essential for basic metabolism and survival.
7) Assertion (A): Secondary metabolites are not directly involved in plant growth.
Reason (R): They play important roles in defense and ecological interactions.
(1) Both A and R are true, R explains A
(2) Both A and R are true, R does not explain A
(3) A is true, R is false
(4) A is false, R is true
Explanation: The correct answer is (1) Both A and R are true, R explains A. Secondary metabolites like alkaloids and phenolics are non-essential for growth but important for plant protection, competition, and signaling. Thus, their ecological roles explain why they are not directly involved in growth.
8) Match the following plant metabolites with their examples:
A. Alkaloid - (i) Morphine
B. Terpenoid - (ii) Rubber
C. Phenolic - (iii) Curcumin
D. Glycoside - (iv) Digitoxin
Options:
(1) A-i, B-ii, C-iii, D-iv
(2) A-ii, B-i, C-iv, D-iii
(3) A-iii, B-iv, C-i, D-ii
(4) A-i, B-iv, C-ii, D-iii
Explanation: The correct answer is (1) A-i, B-ii, C-iii, D-iv. Alkaloids like morphine, terpenoids like rubber, phenolics like curcumin, and glycosides like digitoxin represent major secondary metabolite classes with distinct functions ranging from medicinal to ecological defense.
9) Fill in the blank: ________ is an anticancer alkaloid obtained from Catharanthus roseus.
(1) Morphine
(2) Vinblastine
(3) Curcumin
(4) Rubber
Explanation: The correct answer is (2) Vinblastine. It is a secondary metabolite alkaloid used as a chemotherapeutic drug derived from the Madagascar periwinkle. Unlike morphine or curcumin, vinblastine specifically inhibits cell division and is used in treating cancers like Hodgkin’s lymphoma and leukemia.
10) Choose the correct statements:
A. Primary metabolites include glucose and amino acids.
B. Secondary metabolites are essential for growth.
C. Morphine and curcumin are secondary metabolites.
D. Rubber is a terpenoid.
Options:
(1) A and B
(2) B and C
(3) A, C and D
(4) A and D
Explanation: The correct answer is (3) A, C and D. Primary metabolites like glucose and amino acids are essential. Secondary metabolites like morphine and curcumin play ecological roles. Rubber, a terpenoid, is also a secondary metabolite. Statement B is incorrect because secondary metabolites are not essential for growth.
Topic: DNA Replication
Subtopic: Replication Rate in Prokaryotes
Keyword Definitions:
E. coli: A common bacterium used as a model organism in molecular biology studies.
Base pairs: Paired nucleotides in DNA (A-T, G-C) forming the double helix.
Replication: Process of copying DNA to produce two identical DNA molecules.
Polymerization: Formation of a DNA strand by linking nucleotides via DNA polymerase.
Replication rate: Speed at which DNA polymerase adds nucleotides per second.
Prokaryotes: Organisms without a nucleus; DNA replication occurs in cytoplasm.
Double helix: Structure of DNA formed by two complementary strands.
DNA polymerase: Enzyme catalyzing addition of nucleotides during DNA replication.
Lead Question - 2020 (COVID Reexam)
E.coli has only 4.6 × 106 base pairs and completes the process of replication within 18 minutes; then the average rate of polymerization is approximate-
1. 2000 base pairs/second
2. 3000 base pairs/second
3. 4000 base pairs/second
4. 1000 base pairs/second
Explanation: E. coli has 4.6 × 106 base pairs. Replication time is 18 minutes = 1080 seconds. Average rate = total base pairs / time = 4.6 × 106 / 1080 ≈ 4259 ≈ 4000 base pairs/second. Correct answer: Option 3.
1. Single Correct Answer MCQ:
Which enzyme is responsible for adding nucleotides during replication?
1. DNA ligase
2. DNA polymerase
3. Helicase
4. Topoisomerase
Explanation: DNA polymerase catalyzes the addition of nucleotides to the growing DNA strand during replication, ensuring accurate synthesis. Helicase unwinds DNA, ligase joins fragments, and topoisomerase relieves supercoiling. Correct answer: Option 2.
2. Single Correct Answer MCQ:
The replication of E. coli DNA is completed in approximately:
1. 18 seconds
2. 18 minutes
3. 18 hours
4. 180 minutes
Explanation: E. coli completes DNA replication in 18 minutes, demonstrating rapid polymerization in prokaryotes. This short time is due to circular DNA and multiple replication forks. Answer: Option 2.
3. Single Correct Answer MCQ:
What is the total number of base pairs in E. coli genome?
1. 4.6 × 103
2. 4.6 × 106
3. 4.6 × 109
4. 46
Explanation: The E. coli genome contains approximately 4.6 × 106 base pairs forming a single circular DNA molecule. This small genome allows rapid replication. Correct answer: Option 2.
4. Single Correct Answer MCQ:
The time taken to replicate 1 base pair at the average rate is approximately:
1. 0.25 milliseconds
2. 0.25 seconds
3. 2.5 microseconds
4. 1 second
Explanation: At 4000 base pairs/second, time per base pair = 1/4000 sec = 0.00025 sec = 0.25 milliseconds. This shows the high efficiency of E. coli replication machinery. Correct answer: Option 1.
5. Single Correct Answer MCQ:
Which structure allows faster replication in prokaryotes?
1. Linear DNA
2. Circular DNA
3. Mitochondrial DNA
4. Plasmid only
Explanation: Circular DNA in prokaryotes like E. coli allows bidirectional replication from a single origin, leading to faster replication compared to linear DNA. Plasmids replicate independently but are smaller. Answer: Option 2.
6. Single Correct Answer MCQ:
What is the main reason E. coli replicates DNA rapidly?
1. Multiple chromosomes
2. Multiple replication origins
3. Single circular chromosome
4. Lack of DNA polymerase
Explanation: E. coli has a single circular chromosome with a single origin of replication, but replication is bidirectional, allowing rapid completion in 18 minutes. This efficiency supports fast bacterial growth. Correct answer: Option 3.
7. Assertion-Reason MCQ:
Assertion (A): E. coli completes replication in 18 minutes.
Reason (R): Its genome is small and circular, allowing fast bidirectional replication.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true
Explanation: E. coli replicates its 4.6 × 106 base pair circular genome in 18 minutes due to bidirectional replication from a single origin. Both assertion and reason are correct, and reason explains assertion accurately. Answer: Option 1.
8. Matching Type MCQ:
Column I Column II
(a) DNA polymerase (i) Unwinds DNA
(b) Helicase (ii) Adds nucleotides
(c) Ligase (iii) Joins Okazaki fragments
(d) Topoisomerase (iv) Relieves supercoiling
1. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
3. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
4. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Explanation: DNA polymerase adds nucleotides, helicase unwinds DNA, ligase joins Okazaki fragments, and topoisomerase relieves supercoiling. Correct match: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv). Answer: Option 1.
9. Fill in the blanks:
The bidirectional replication in E. coli begins at a single ______ of replication.
1. Origin
2. Terminus
3. Fork
4. Centromere
Explanation: E. coli DNA replication begins at a single origin of replication, proceeding bidirectionally around the circular chromosome until termination. This strategy allows rapid genome duplication. Answer: Option 1.
10. Choose the correct statements MCQ:
Select the correct statements about E. coli DNA replication:
(a) Replication is bidirectional
(b) Rate ≈ 4000 base pairs
Copic: DNA Structure and Function
Subtopic: Polynucleotide Chain and Nitrogenous Bases
Keyword Definitions:
Polynucleotide chain: A polymer of nucleotides linked by phosphodiester bonds forming DNA or RNA strands.
Nitrogenous base: Organic molecules (A, T, G, C) attached to the sugar in nucleotides.
Pentose sugar: Five-carbon sugar in nucleotides; deoxyribose in DNA, ribose in RNA.
1'C of sugar: Carbon atom in sugar where nitrogenous base attaches.
3'C and 5'C: Carbon atoms in sugar forming phosphodiester linkages between nucleotides.
Phosphodiester bond: Covalent bond linking the 3'C of one sugar to 5'C of another.
DNA strand: Sequence of nucleotides forming genetic material.
OH group: Hydroxyl group on sugar that forms bond with phosphate or base.
Lead Question - 2020 (COVID Reexam)
In the polynucleotide chain of DNA, a nitrogenous base is linked to the –OH of:
1. 2'C pentose sugar
2. 3'C pentose sugar
3. 5'C pentose sugar
4. 1'C pentose sugar
Explanation: In DNA, each nitrogenous base (A, T, G, C) is covalently attached to the 1' carbon of the deoxyribose sugar via a β-glycosidic bond. This linkage forms the nucleoside, which combines with phosphate groups to create the polynucleotide chain. Correct answer: Option 4.
1. Single Correct Answer MCQ:
Which carbon of sugar forms the phosphodiester bond with another nucleotide?
1. 1'C
2. 2'C
3. 3'C
4. 5'C
Explanation: Phosphodiester bonds connect the 3' hydroxyl group of one sugar to the 5' phosphate of the next nucleotide, forming the backbone of DNA. The 1'C attaches to the base. Answer: Option 3.
2. Single Correct Answer MCQ:
Which sugar is present in DNA nucleotides?
1. Ribose
2. Deoxyribose
3. Glucose
4. Fructose
Explanation: DNA contains deoxyribose sugar, lacking one oxygen at the 2' carbon compared to ribose in RNA. Glucose and fructose are monosaccharides but not part of nucleotides. Correct answer: Option 2.
3. Single Correct Answer MCQ:
Which base is complementary to adenine in DNA?
1. Guanine
2. Cytosine
3. Thymine
4. Uracil
Explanation: In DNA, adenine forms two hydrogen bonds with thymine, maintaining base pairing specificity. Cytosine pairs with guanine, and uracil is in RNA. Answer: Option 3.
4. Single Correct Answer MCQ:
A nucleotide consists of:
1. Sugar only
2. Base and phosphate only
3. Sugar and base only
4. Sugar, base, and phosphate
Explanation: A nucleotide is composed of a pentose sugar, a nitrogenous base, and a phosphate group. Together, nucleotides polymerize to form DNA or RNA chains. Answer: Option 4.
5. Single Correct Answer MCQ:
Which group on sugar links with phosphate in DNA backbone?
1. 1'OH
2. 2'OH
3. 3'OH
4. 5'OH
Explanation: The 3' hydroxyl of sugar forms phosphodiester bonds with 5' phosphate of the next nucleotide, creating DNA's sugar-phosphate backbone. The 1' carbon links to the nitrogenous base. Answer: Option 3.
6. Single Correct Answer MCQ:
In DNA, thymine is attached to which carbon of sugar?
1. 1'C
2. 2'C
3. 3'C
4. 5'C
Explanation: Nitrogenous bases in DNA, including thymine, attach covalently to the 1' carbon of deoxyribose sugar, forming a nucleoside. This allows polymerization through phosphodiester bonds on 3' and 5' carbons. Answer: Option 1.
7. Assertion-Reason MCQ:
Assertion (A): Nitrogenous base is linked to 1' carbon of sugar.
Reason (R): The base forms a glycosidic bond with sugar in nucleotides.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true
Explanation: The nitrogenous base attaches to the 1' carbon of the sugar via a β-glycosidic bond in nucleotides. Both assertion and reason are correct, and the reason accurately explains the assertion. Answer: Option 1.
8. Matching Type MCQ:
Column I Column II
(a) 1' carbon (i) Base attachment
(b) 3' carbon (ii) Phosphodiester bond
(c) 5' carbon (iii) Terminal phosphate
(d) Nitrogenous base (iv) Purine or pyrimidine
1. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
2. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
3. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
4. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Explanation: The 1' carbon attaches to the base, 3' carbon forms phosphodiester bonds, 5' carbon carries terminal phosphate, and nitrogenous base is purine or pyrimidine. Correct match: (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv). Answer: Option 1.
9. Fill in the blanks:
In DNA, a nitrogenous base is covalently attached to the ______ carbon of deoxyribose sugar.
1. 1'
2. 2'
3. 3'
4. 5'
Explanation: The nitrogenous base in DNA is attached to the 1' carbon of deoxyribose sugar via a β-glycosidic bond forming a nucleoside.
Subtopic: Discovery of Nucleic Acids
Keyword Definitions:
Nuclein: The substance discovered in the nucleus, later known as DNA, first isolated by Friedrich Miescher.
DNA: Deoxyribonucleic acid, the molecule carrying genetic information in cells.
Friedrich Miescher: Swiss biochemist who first isolated nuclein from pus cells in 1869.
Chargaff: Discovered base pairing rules (A=T, G≡C) in DNA.
Franklin: Rosalind Franklin obtained X-ray diffraction images of DNA, crucial for structure discovery.
Mendel: Father of genetics, studied inheritance in pea plants but did not study DNA.
Genetic material: Molecules responsible for storing and transmitting hereditary information.
X-ray diffraction: Technique to study molecular structure used by Franklin for DNA.
Lead Question - 2020 (COVID Reexam)
The term 'Nuclein' for the genetic material was used by:
1. Franklin
2. Meischer
3. Chargaff
4. Mendel
Explanation: Friedrich Miescher first isolated a substance from the nucleus in 1869, which he called 'Nuclein'. This substance was later identified as DNA, the carrier of genetic information. Franklin, Chargaff, and Mendel contributed to DNA structure or genetics, but the term 'Nuclein' was coined by Miescher. Correct answer: Option 2.
1. Single Correct Answer MCQ:
Who discovered DNA base pairing rules?
1. Franklin
2. Miescher
3. Chargaff
4. Mendel
Explanation: Erwin Chargaff discovered that in DNA, adenine equals thymine and guanine equals cytosine, establishing base pairing rules essential for DNA structure. Franklin obtained X-ray diffraction images, Miescher discovered nuclein, and Mendel studied inheritance in pea plants. Answer: Option 3.
2. Single Correct Answer MCQ:
Which scientist used X-ray diffraction to study DNA?
1. Mendel
2. Franklin
3. Miescher
4. Chargaff
Explanation: Rosalind Franklin used X-ray diffraction to reveal DNA's helical structure, providing critical data for Watson and Crick's double helix model. Miescher isolated nuclein, Chargaff discovered base ratios, and Mendel worked on inheritance patterns. Answer: Option 2.
3. Single Correct Answer MCQ:
Who first isolated nuclein from cell nuclei?
1. Franklin
2. Miescher
3. Chargaff
4. Mendel
Explanation: Friedrich Miescher in 1869 isolated nuclein from the nuclei of pus cells. This nuclein was later identified as DNA. Franklin studied DNA structure, Chargaff discovered base ratios, and Mendel discovered inheritance laws. Answer: Option 2.
4. Single Correct Answer MCQ:
Nuclein later became known as:
1. RNA
2. Protein
3. DNA
4. Chromatin
Explanation: Nuclein isolated by Miescher was later identified as DNA, the hereditary material carrying genetic information. RNA, protein, and chromatin are cellular components but nuclein specifically refers to DNA. Answer: Option 3.
5. Single Correct Answer MCQ:
Which scientist studied pea plants and inheritance?
1. Miescher
2. Franklin
3. Mendel
4. Chargaff
Explanation: Gregor Mendel studied inheritance in pea plants, formulating laws of segregation and independent assortment. He did not study DNA. Miescher isolated nuclein, Franklin studied DNA structure, and Chargaff discovered base ratios. Answer: Option 3.
6. Single Correct Answer MCQ:
Which component is the genetic material in cells?
1. RNA
2. DNA
3. Protein
4. Lipid
Explanation: DNA is the genetic material in most cells, carrying hereditary information. RNA plays roles in protein synthesis, proteins perform structural and enzymatic functions, and lipids form membranes. Nuclein was an early term for DNA. Answer: Option 2.
7. Assertion-Reason MCQ:
Assertion (A): Miescher discovered nuclein in 1869.
Reason (R): Nuclein is composed of DNA and protein found in the nucleus.
1. Both A and R true, R correct explanation
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true
Explanation: Miescher discovered nuclein from nuclei, which contained DNA and some protein. Both assertion and reason are correct, and reason accurately explains the assertion. Answer: Option 1.
8. Matching Type MCQ:
Column I Column II
(a) Miescher (i) Base ratios in DNA
(b) Franklin (ii) X-ray diffraction
(c) Chargaff (iii) Isolated nuclein
(d) Mendel (iv) Pea plant inheritance
1. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
2. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
3. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
4. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Explanation: Miescher isolated nuclein, Franklin studied DNA by X-ray diffraction, Chargaff discovered base ratios, Mendel studied inheritance. Correct matching: (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv). Answer: Option 1.
9. Fill in the blanks:
The Swiss scientist who first coined the term 'Nuclein' is ________.
1. Franklin
2. Miescher
3. Chargaff
4. Mendel
Explanation: Friedrich Miescher, a Swiss biochemist, first isolated nuclein from cell nuclei and coined the term. This nuclein was later recognized as DNA, the carrier of genetic information. Franklin, Chargaff, and Mendel contributed to DNA structure or inheritance but did not name nuclein. Answer: Option 2.
10. Choose correct statements:
(a) Miescher isolated nuclein
(b) Nuclein is composed of DNA and protein
(c) Franklin discovered nuclein
(d) Mendel studied inheritance in pea plants
1. a, b, d only
2. a and c only
3. b and c only
4. All statements correct
Explanation: Miescher isolated nuclein containing DNA and protein. Mendel studied pea plant inheritance. Franklin did not discover nuclein but studied DNA structure. Therefore, correct statements are a, b, d only. Answer: Option 1.
Topic: Proteins, Carbohydrates, and Plant Hormones
Subtopic: Structure and Function of Biological Molecules
Keyword Definitions:
Aquaporin: Integral membrane protein that facilitates water transport across membranes.
Asparagine: Amino acid containing an amide functional group in its side chain.
Abscisic acid (ABA): Plant hormone, belongs to carotenoids, involved in stress responses and seed dormancy.
Chitin: Structural polysaccharide found in fungal cell walls and arthropod exoskeletons.
Amide: Organic functional group containing -CONH2.
Polysaccharide: Polymer of monosaccharides forming complex carbohydrates.
Polypeptide: Chain of amino acids forming proteins.
Carotenoids: Pigments involved in photosynthesis and photoprotection in plants.
Lead Question - 2020 (COVID Reexam)
Match the following :
(a) Aquaporin (i) Amide
(b) Asparagine (ii) Polysaccharide
(c) Abscisic acid (iii) Polypeptide
(d) Chitin (iv) Carotenoids
Select the correct option :
1. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
2. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
3. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
4. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Explanation: Aquaporin is a polypeptide forming water channels in membranes. Asparagine contains an amide group in its side chain. Abscisic acid is a carotenoid-derived plant hormone, and chitin is a polysaccharide. Therefore, the correct matching is option 1: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii), showing proper classification of biomolecules. (50 words)
Guessed Question 1. Aquaporins are classified as :
1. Lipids
2. Proteins
3. Polysaccharides
4. Nucleic acids
Explanation: Aquaporins are integral membrane proteins forming channels for water transport across biological membranes. They are not lipids, polysaccharides, or nucleic acids. They are crucial for water balance in cells and tissues. Therefore, the correct answer is option 2: Proteins. (50 words)
Guessed Question 2. Asparagine is an example of :
1. Aromatic amino acid
2. Amide-containing amino acid
3. Sulfur-containing amino acid
4. Non-polar amino acid
Explanation: Asparagine contains a side chain with an amide functional group (-CONH2). It is polar but uncharged, not aromatic or sulfur-containing. This property is important in protein folding and hydrogen bonding. Therefore, the correct answer is option 2: Amide-containing amino acid. (50 words)
Guessed Question 3. Abscisic acid is derived from :
1. Proteins
2. Polysaccharides
3. Carotenoids
4. Lipids
Explanation: Abscisic acid (ABA) is a plant hormone derived from carotenoid pigments. It regulates stress responses, seed dormancy, and stomatal closure. It is not derived from proteins, polysaccharides, or lipids directly. Correct answer is option 3: Carotenoids. (50 words)
Guessed Question 4. Chitin is a :
1. Protein
2. Polysaccharide
3. Lipid
4. Nucleic acid
Explanation: Chitin is a structural polysaccharide composed of N-acetylglucosamine units. It is found in fungal cell walls and arthropod exoskeletons. It is not a protein, lipid, or nucleic acid. Its rigidity provides structural support. Therefore, the correct answer is option 2: Polysaccharide. (50 words)
Guessed Question 5. Assertion-Reason Question:
Assertion (A): Aquaporins are made of polypeptides.
Reason (R): Polypeptides are polymers of amino acids forming proteins.
1. Both A and R true, R explains A
2. Both A and R true, R not correct explanation
3. A true, R false
4. A false, R true
Explanation: Aquaporins are membrane proteins composed of polypeptide chains. Polypeptides are indeed polymers of amino acids forming proteins. Both assertion and reason are true, and reason correctly explains the assertion. Therefore, correct answer is option 1. Polypeptide structure enables selective water transport in aquaporins. (50 words)
Guessed Question 6. Match the following:
(a) Cellulose (i) Structural polysaccharide
(b) Collagen (ii) Protein
(c) Starch (iii) Energy storage polysaccharide
(d) Hemoglobin (iv) Oxygen transport protein
1. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
2. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
3. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
4. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Explanation: Cellulose is a structural polysaccharide in plants. Collagen is a structural protein in connective tissues. Starch is a polysaccharide for energy storage. Hemoglobin is a protein for oxygen transport. Correct matching is option 1: (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv). This reflects biological molecule functions. (50 words)
Guessed Question 7. Fill in the blank:
Abscisic acid helps in ______ of seeds.
1. Germination
2. Dormancy
3. Growth
4. Photosynthesis
Explanation: Abscisic acid induces seed dormancy and inhibits premature germination. It helps seeds survive adverse conditions by controlling water loss and metabolic activity. It does not directly promote germination, growth, or photosynthesis. Therefore, the correct answer is option 2: Dormancy. (50 words)
Guessed Question 8. Choose correct statements:
1. Aquaporins are polypeptides.
2. Chitin is a polysaccharide.
3. Abscisic acid is a lipid.
4. Asparagine contains an amide group.
Options:
A. 1 and 2
B. 1, 2 and 4
C. 2 and 3
D. All 1, 2, 3, 4
Explanation: Aquaporins are polypeptides, chitin is a polysaccharide, and asparagine contains an amide group. Abscisic acid is derived from carotenoids, not a general lipid. Therefore, correct statements are 1, 2, and 4. The correct answer is option B. This demonstrates accurate classification of biomolecules. (50 words)
Topic: Amino Acids and Lipids
Subtopic: Structure and Functions
Keyword Definitions:
Amino acids: Organic molecules with amino and carboxyl groups forming proteins.
Sulphur: An essential element present in certain amino acids like cysteine and methionine.
Lipids: Hydrophobic molecules including fats, oils, phospholipids, and steroids.
Lecithin: A phospholipid containing choline, phosphate, and fatty acids.
Aromatic amino acids: Amino acids like tyrosine, tryptophan, and phenylalanine with aromatic rings.
Lead Question - 2020 (COVID Reexam)
Identify the statement which is incorrect.
1. Sulphur is an integral part of cysteine.
2. Glycine is an example of lipids.
3. Lecithin contains a phosphorus atom in its structure.
4. Tyrosine possesses an aromatic ring in its structure.
Explanation: The incorrect statement is option 2. Glycine is the simplest amino acid, not a lipid. Sulphur is present in cysteine, lecithin has phosphorus, and tyrosine has an aromatic ring. Amino acids are protein building blocks, while lipids are separate biomolecules. Correct answer is Glycine is not lipid. (50 words)
1. Which amino acid contains sulphur?
1. Valine
2. Cysteine
3. Glycine
4. Alanine
Explanation: The amino acid cysteine contains sulphur in its side chain. Sulphur helps in disulfide bond formation, providing structural stability to proteins. Other amino acids like glycine, valine, and alanine do not contain sulphur. Thus, cysteine plays a key role in protein folding and function. Correct answer is cysteine. (50 words)
2. Glycine is classified as:
1. Non-polar amino acid
2. Simplest amino acid
3. Essential amino acid
4. Aromatic amino acid
Explanation: Glycine is the simplest amino acid with a single hydrogen atom as its side chain. It is a non-essential amino acid and not aromatic. It contributes flexibility to protein structures due to its small size. Correct classification of glycine is the simplest amino acid. Correct answer is option 2. (50 words)
3. Lecithin belongs to which biomolecule category?
1. Steroids
2. Phospholipids
3. Proteins
4. Amino acids
Explanation: Lecithin is a phospholipid commonly found in biological membranes. It contains fatty acids, phosphate, and choline groups. It helps in maintaining membrane fluidity and also acts as an emulsifier in food industries. It is not a protein or amino acid. The correct answer is phospholipids. (50 words)
4. Which amino acid has an aromatic ring?
1. Tyrosine
2. Glycine
3. Serine
4. Threonine
Explanation: Tyrosine contains a phenolic aromatic ring in its side chain. This makes it an aromatic amino acid, along with tryptophan and phenylalanine. Aromatic amino acids absorb UV light and play important roles in protein structure and enzyme activity. Correct answer is Tyrosine. (50 words)
5. Assertion-Reason Question:
Assertion (A): Cysteine can form disulfide bonds.
Reason (R): Glycine has a sulfur atom in its structure.
1. Both A and R are true, R is correct explanation.
2. Both A and R are true, R is not correct explanation.
3. A is true, R is false.
4. A is false, R is true.
Explanation: Cysteine forms disulfide bonds due to its thiol (-SH) group, stabilizing proteins. Glycine does not contain sulfur, so the reason is false. Therefore, A is true and R is false. This distinction is essential in protein chemistry. Correct answer is option 3. (50 words)
6. Matching Type Question:
Match the following:
(a) Phenylalanine (i) Phospholipid
(b) Lecithin (ii) Aromatic amino acid
(c) Methionine (iii) Sulfur-containing amino acid
(d) Cholesterol (iv) Steroid
1. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
3. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
4. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Explanation: Phenylalanine is an aromatic amino acid. Lecithin is a phospholipid. Methionine is a sulfur-containing amino acid. Cholesterol is a steroid. Thus, correct matching is (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv). This classification highlights major biomolecules and their importance in structure and metabolism. Correct answer is option 1. (50 words)
7. Fill in the Blank:
The simplest amino acid with hydrogen as side chain is ________.
1. Alanine
2. Glycine
3. Valine
4. Proline
Explanation: Glycine has hydrogen as its side chain, making it the simplest amino acid. Its small size provides flexibility to proteins and helps in the formation of tight turns. Other amino acids have bulkier side chains. Therefore, the correct answer is glycine. Correct answer is option 2. (50 words)
8. Choose the correct statements:
1. Lecithin is a protein.
2. Tyrosine has an aromatic ring.
3. Glycine is a lipid.
4. Cysteine contains sulphur.
Options:
A. 1 and 2
B. 2 and 3
C. 2 and 4
D. 1 and 3
Explanation: Correct statements are that Tyrosine has an aromatic ring and Cysteine contains sulphur. Lecithin is not a protein but a phospholipid, and Glycine is an amino acid not lipid. Therefore, the right choice is option C (2 and 4). Correct answer is option C. (50 words)
Keyword Definitions:
Glycosidic bond – Covalent bond linking carbohydrate monomers in polysaccharides.
Peptide bond – Covalent bond linking amino acids in proteins or peptides.
Cellulose – Polysaccharide of β-glucose units with glycosidic linkages, structural component of plant cell wall.
Inulin – Polysaccharide composed of fructose units, used as storage carbohydrate in plants.
Insulin – Peptide hormone regulating blood glucose, composed of amino acids linked by peptide bonds.
Chitin – Polysaccharide forming fungal cell walls and exoskeletons of arthropods.
Trypsin – Digestive protease enzyme composed of amino acids linked via peptide bonds.
Lead Question - 2020
Identify the substances having glycosidic bond and peptide bond, respectively in their structure:
(1) Cellulose, lecithin
(2) Inulin, insulin
(3) Chitin, cholesterol
(4) Glycerol, trypsin
Explanation: Inulin is a polysaccharide containing glycosidic bonds, while insulin is a peptide hormone composed of amino acids connected by peptide bonds. This combination correctly represents glycosidic and peptide bonds. Correct answer is (2) Inulin, insulin.
1. Single Correct Answer: Which of the following contains glycosidic bonds?
(1) Glucose
(2) Starch
(3) Alanine
(4) Urea
Explanation: Starch is a polysaccharide made of glucose units linked by glycosidic bonds. Correct answer is (2) Starch.
2. Single Correct Answer: The bond linking amino acids in proteins is:
(1) Glycosidic
(2) Hydrogen
(3) Peptide
(4) Phosphodiester
Explanation: Peptide bonds are covalent bonds formed between amino acids in proteins, essential for polypeptide chains. Correct answer is (3) Peptide.
3. Single Correct Answer: Chitin is composed of:
(1) Amino acids
(2) Fructose units
(3) N-acetylglucosamine units
(4) Glycerol
Explanation: Chitin is a polysaccharide of N-acetylglucosamine linked by glycosidic bonds forming fungal cell walls and arthropod exoskeletons. Correct answer is (3) N-acetylglucosamine units.
4. Assertion-Reason:
Assertion (A): Insulin is a protein.
Reason (R): It contains amino acids linked by peptide bonds.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Insulin is a peptide hormone composed of amino acids joined by peptide bonds, making both the assertion and reason correct and reason explains assertion. Correct answer is (1).
5. Single Correct Answer: Which polysaccharide is used as a storage carbohydrate in plants?
(1) Insulin
(2) Inulin
(3) Collagen
(4) Trypsin
Explanation: Inulin is a plant storage polysaccharide composed of fructose units linked by glycosidic bonds. Correct answer is (2) Inulin.
6. Single Correct Answer: The peptide hormone regulating blood sugar is:
(1) Glucagon
(2) Insulin
(3) Starch
(4) Chitin
Explanation: Insulin is a peptide hormone that regulates blood glucose levels, consisting of amino acids linked by peptide bonds. Correct answer is (2) Insulin.
7. Matching Type: Match Column I with Column II:
a. Polysaccharide – i. Inulin
b. Protein hormone – ii. Insulin
c. Structural polysaccharide – iii. Chitin
d. Enzyme – iv. Trypsin
(1) a-i, b-ii, c-iii, d-iv
(2) a-iii, b-i, c-iv, d-ii
(3) a-iv, b-iii, c-ii, d-i
(4) a-ii, b-iv, c-i, d-iii
Explanation: Inulin is a polysaccharide (a-i), insulin is a protein hormone (b-ii), chitin is a structural polysaccharide (c-iii), and trypsin is an enzyme (d-iv). Correct answer is (1).
8. Fill in the blank: _______ is a polysaccharide composed of fructose units linked by glycosidic bonds.
(1) Inulin
(2) Insulin
(3) Chitin
(4) Collagen
Explanation: Inulin is a polysaccharide of fructose units connected via glycosidic bonds and serves as a storage carbohydrate in plants. Correct answer is (1) Inulin.
9. Single Correct Answer: Which of the following contains peptide bonds?
(1) Inulin
(2) Insulin
(3) Cellulose
(4) Chitin
Explanation: Insulin is a peptide hormone composed of amino acids linked by peptide bonds, forming its functional structure. Correct answer is (2) Insulin.
10. Choose the correct statements:
(a) Glycosidic bonds link monosaccharides
(b) Peptide bonds link amino acids
(c) Inulin is a polysaccharide
(d) Insulin is a protein hormone
(1) a, b, c, d
(2) a, c, d
(3) b, d
(4) a, b, c
Explanation: All four statements are correct: glycosidic bonds link monosaccharides (a), peptide bonds link amino acids (b), inulin is a polysaccharide (c), and insulin is a protein hormone (d). Correct answer is (1) a, b, c, d.
Keyword Definitions:
Inhibitor of catalytic activity – A molecule that decreases or stops the activity of an enzyme, preventing substrate conversion to product.
Peptide bonds – Covalent bonds linking amino acids in proteins or polypeptides.
Chitin – Structural polysaccharide forming the cell wall in fungi and exoskeletons of arthropods.
Secondary metabolite – Organic compounds not directly involved in normal growth, development, or reproduction, often involved in defense, e.g., toxins.
Ricin – Toxic protein from castor bean that inhibits protein synthesis.
Malonate – Competitive inhibitor of succinate dehydrogenase in the citric acid cycle.
Collagen – Structural protein abundant in connective tissues, providing strength and support.
Lead Question - 2020
Match the following:
(a) Inhibitor of catalytic activity
(b) Possess peptide bonds
(c) Cell wall material in fungi
(d) Secondary metabolite
(i) Ricin
(ii) Malonate
(iii) Chitin
(iv) Collagen
Choose the correct option:
(A) a-iii, b-iv, c-i, d-ii
(B) a-ii, b-iii, c-i, d-iv
(C) a-ii, b-iv, c-iii, d-i
(D) a-iii, b-i, c-iv, d-ii
Explanation: Malonate is an inhibitor of catalytic activity (a-ii). Collagen contains peptide bonds (b-iv). Chitin forms the cell wall in fungi (c-iii). Ricin is a secondary metabolite (d-i) acting as a toxin. Correct answer is (C) a-ii, b-iv, c-iii, d-i.
1. Single Correct Answer: Which of the following is a competitive inhibitor of succinate dehydrogenase?
(1) Ricin
(2) Malonate
(3) Chitin
(4) Collagen
Explanation: Malonate competitively inhibits succinate dehydrogenase in the citric acid cycle, blocking substrate binding. Correct answer is (2) Malonate.
2. Single Correct Answer: Proteins are characterized by which bond linking amino acids?
(1) Glycosidic bonds
(2) Phosphodiester bonds
(3) Peptide bonds
(4) Hydrogen bonds
Explanation: Peptide bonds are covalent links between amino acids in proteins, essential for polypeptide formation. Correct answer is (3) Peptide bonds.
3. Single Correct Answer: The structural polysaccharide forming fungal cell walls is:
(1) Cellulose
(2) Chitin
(3) Starch
(4) Glycogen
Explanation: Chitin is a polymer of N-acetylglucosamine forming rigid fungal cell walls. Correct answer is (2) Chitin.
4. Assertion-Reason:
Assertion (A): Ricin is a toxic protein.
Reason (R): Ricin inhibits protein synthesis in target cells.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Ricin is a secondary metabolite toxin that inhibits protein synthesis, making both statements correct and the reason explains the assertion. Correct answer is (1).
5. Single Correct Answer: Collagen is an example of:
(1) Structural protein
(2) Enzyme inhibitor
(3) Secondary metabolite
(4) Polysaccharide
Explanation: Collagen is a structural protein in connective tissue, rich in peptide bonds, providing tensile strength. Correct answer is (1) Structural protein.
6. Single Correct Answer: Secondary metabolites are primarily involved in:
(1) Growth and reproduction
(2) Defense and competition
(3) Energy production
(4) Protein synthesis
Explanation: Secondary metabolites, like ricin, are not required for growth but help in defense, interspecies competition, and ecological interactions. Correct answer is (2) Defense and competition.
7. Matching Type: Match Column I with Column II:
a. Enzyme inhibitor – i. Malonate
b. Structural protein – ii. Collagen
c. Fungal cell wall – iii. Chitin
d. Toxic secondary metabolite – iv. Ricin
(1) a-i, b-ii, c-iii, d-iv
(2) a-iv, b-i, c-ii, d-iii
(3) a-iii, b-iv, c-i, d-ii
(4) a-ii, b-iii, c-iv, d-i
Explanation: Malonate inhibits enzymes (a-i), collagen is a structural protein (b-ii), chitin forms fungal cell walls (c-iii), and ricin is a toxic secondary metabolite (d-iv). Correct answer is (1).
8. Fill in the blank: _______ is a polymer forming fungal cell walls.
(1) Chitin
(2) Collagen
(3) Malonate
(4) Ricin
Explanation: Chitin, a long-chain polymer of N-acetylglucosamine, provides rigidity to fungal cell walls. Correct answer is (1) Chitin.
9. Single Correct Answer: Which compound inhibits succinate dehydrogenase competitively?
(1) Ricin
(2) Malonate
(3) Chitin
(4) Collagen
Explanation: Malonate acts as a competitive inhibitor of succinate dehydrogenase, blocking substrate access to the enzyme's active site. Correct answer is (2) Malonate.
10. Choose the correct statements:
(a) Collagen contains peptide bonds
(b) Chitin is structural polysaccharide in fungi
(c) Ricin is a secondary metabolite toxin
(d) Malonate is an enzyme inhibitor
(1) a, b, c, d
(2) a, b, d
(3) b, c, d
(4) a, c, d
Explanation: All four statements are correct: collagen has peptide bonds (a), chitin is structural (b), ricin is a secondary metabolite toxin (c), and malonate is an enzyme inhibitor (d). Correct answer is (1) a, b, c, d.
Keyword Definitions:
Translation – Process by which mRNA is decoded by ribosomes to synthesize proteins.
tRNA – Transfer RNA, brings specific amino acids to the ribosome during translation.
Aminoacylation – Attachment of an amino acid to its corresponding tRNA by aminoacyl-tRNA synthetase.
Anti-codon – Three-nucleotide sequence on tRNA complementary to codon on mRNA.
mRNA – Messenger RNA, carries genetic code from DNA to ribosome for protein synthesis.
Ribosome – Cellular machinery that reads mRNA and assembles amino acids into a polypeptide chain.
Lead Question - 2020
The first phase of translation is:
(1) Aminoacylation of tRNA
(2) Recognition of an anti-codon
(3) Binding of mRNA to ribosome
(4) Recognition of DNA molecule
Explanation: Translation begins with aminoacylation of tRNA, which attaches a specific amino acid to its tRNA via aminoacyl-tRNA synthetase. This step is essential before codon-anticodon recognition and ribosome binding. DNA recognition is irrelevant at this stage. Correct answer is (1) Aminoacylation of tRNA.
1. Single Correct Answer: The codon-anticodon pairing occurs at:
(1) Ribosome A site
(2) Ribosome P site
(3) Ribosome E site
(4) Nucleus
Explanation: During translation, codon-anticodon recognition occurs at the ribosome A site where tRNA binds complementary to mRNA codon. Correct answer is (1).
2. Single Correct Answer: The enzyme responsible for aminoacylation is:
(1) RNA polymerase
(2) Aminoacyl-tRNA synthetase
(3) Ligase
(4) Helicase
Explanation: Aminoacyl-tRNA synthetase catalyzes the attachment of specific amino acids to their tRNAs, forming aminoacyl-tRNA, which is the first step in translation. Correct answer is (2).
3. Single Correct Answer: Start codon on mRNA is:
(1) UAG
(2) AUG
(3) UAA
(4) UGA
Explanation: AUG codon on mRNA codes for methionine and acts as the start codon for translation initiation. Correct answer is (2).
4. Assertion (A): Aminoacylation of tRNA is essential before translation.
Reason (R): tRNA without amino acid cannot bind ribosome.
(1) Both A and R true, R explains A
(2) Both A and R true, R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: tRNA must be charged with amino acid before entering translation. Uncharged tRNA cannot deliver amino acids to ribosome, so R explains A. Correct answer is (1).
5. Single Correct Answer: The ribosome binding site on mRNA is called:
(1) Shine-Dalgarno sequence
(2) Poly-A tail
(3) 5’ cap only
(4) Intron
Explanation: The Shine-Dalgarno sequence in prokaryotic mRNA guides ribosome binding at the start codon. Correct answer is (1).
6. Single Correct Answer: The A site of ribosome is responsible for:
(1) Exit of tRNA
(2) Aminoacyl-tRNA entry
(3) Peptide bond formation
(4) mRNA transcription
Explanation: Aminoacyl-tRNA enters the ribosome at A site to pair with codon on mRNA. Correct answer is (2).
7. Matching Type: Match column I with column II:
a. tRNA charging – i. Aminoacylation
b. Start codon – ii. AUG
c. Ribosome site for new tRNA – iii. A site
d. Polypeptide exit – iv. P site
(1) a-i, b-ii, c-iii, d-iv
(2) a-ii, b-i, c-iv, d-iii
(3) a-iii, b-iv, c-ii, d-i
(4) a-iv, b-iii, c-i, d-ii
Explanation: tRNA charging corresponds to aminoacylation (a-i), start codon is AUG (b-ii), A site accepts new tRNA (c-iii), P site holds growing polypeptide (d-iv). Correct answer is (1).
8. Fill in the blank: The first amino acid incorporated during translation in eukaryotes is ______.
(1) Methionine
(2) Formyl-methionine
(3) Leucine
(4) Glycine
Explanation: In eukaryotes, translation begins with methionine incorporated by initiator tRNA. Correct answer is (1).
9. Single Correct Answer: Which molecule carries amino acid to ribosome?
(1) mRNA
(2) tRNA
(3) rRNA
(4) DNA
Explanation: tRNA transports specific amino acids to the ribosome, facilitating translation. Correct answer is (2).
10. Choose the correct statements:
(a) Translation begins with aminoacylation of tRNA
(b) Start codon is AUG
(c) DNA is directly involved in translation
(d) Ribosome has A, P, and E sites
(1) a, b, d only
(2) a, c, d only
(3) b, c, d only
(4) a, b, c only
Explanation: Translation requires tRNA aminoacylation (a), AUG start codon (b), and ribosome sites A, P, E (d). DNA is not directly involved. Correct answer is (1) a, b, d only.
Keyword Definitions:
Floridean starch – Storage polysaccharide found in red algae, similar to glycogen and amylopectin.
Amylopectin – Branched polymer of glucose, component of plant starch.
Glycogen – Branched glucose polymer, energy storage in animals and fungi.
Starch – Plant polysaccharide for energy storage, composed of amylose and amylopectin.
Cellulose – Linear polymer of glucose forming plant cell walls.
Polysaccharide – Long-chain carbohydrate composed of monosaccharide units linked by glycosidic bonds.
Lead Question - 2020
Floridean starch has structure similar to :
(1) Mannitol and algin
(2) Laminarin and cellulose
(3) Starch and cellulose
(4) Amylopectin and glycogen
Explanation: Floridean starch is a branched polysaccharide found in red algae and is structurally similar to amylopectin and glycogen, providing energy storage. Unlike starch in plants, it is more highly branched. Correct answer is (4) Amylopectin and glycogen.
1. Single Correct Answer: Which of the following is the main storage carbohydrate in red algae?
(1) Laminarin
(2) Floridean starch
(3) Starch
(4) Glycogen
Explanation: Red algae store energy mainly in the form of Floridean starch, a branched polymer of glucose similar to glycogen. Correct answer is (2) Floridean starch.
2. Single Correct Answer: Amylopectin is:
(1) Linear glucose polymer
(2) Branched glucose polymer
(3) Fructose polymer
(4) Protein-linked sugar
Explanation: Amylopectin is a branched polymer of glucose forming part of plant starch, providing a compact structure for energy storage. Correct answer is (2) Branched glucose polymer.
3. Single Correct Answer: Glycogen differs from starch by:
(1) Being linear
(2) Being more highly branched
(3) Containing fructose
(4) Being unbranched
Explanation: Glycogen is more highly branched than starch and amylopectin, allowing rapid mobilization of glucose for energy in animals. Correct answer is (2) Being more highly branched.
4. Assertion (A): Floridean starch is similar to glycogen.
Reason (R): It is a highly branched polysaccharide stored in cytoplasm.
(1) Both A and R true and R explains A
(2) Both A and R true but R does not explain A
(3) A true, R false
(4) A false, R true
Explanation: Floridean starch is a highly branched cytoplasmic polysaccharide, similar in structure and function to glycogen. Both Assertion and Reason are true, and R correctly explains A. Correct answer is (1).
5. Single Correct Answer: Starch is composed of:
(1) Amylose only
(2) Amylopectin only
(3) Amylose and amylopectin
(4) Glycogen and amylopectin
Explanation: Starch in plants consists of two glucose polymers, linear amylose and branched amylopectin, providing compact energy storage. Correct answer is (3) Amylose and amylopectin.
6. Single Correct Answer: Laminarin is found in:
(1) Red algae
(2) Brown algae
(3) Green algae
(4) Cyanobacteria
Explanation: Laminarin is a storage polysaccharide in brown algae, consisting of β-glucose units, unlike Floridean starch in red algae. Correct answer is (2) Brown algae.
7. Match the polysaccharide with its source:
a. Floridean starch – i. Red algae
b. Laminarin – ii. Brown algae
c. Starch – iii. Green plants
d. Glycogen – iv. Animals
(1) a-i, b-ii, c-iii, d-iv
(2) a-ii, b-i, c-iii, d-iv
(3) a-iii, b-ii, c-i, d-iv
(4) a-i, b-iii, c-ii, d-iv
Explanation: Floridean starch is found in red algae (a-i), laminarin in brown algae (b-ii), starch in green plants (c-iii), and glycogen in animals (d-iv). Correct answer is (1).
8. Fill in the blank: The storage polysaccharide in red algae is called ______.
(1) Laminarin
(2) Floridean starch
(3) Glycogen
(4) Amylose
Explanation: Red algae store energy in the form of Floridean starch, a branched glucose polymer similar to glycogen. Correct answer is (2) Floridean starch.
9. Single Correct Answer: Cellulose differs from starch in that it:
(1) Is branched
(2) Has α-glucose linkages
(3) Has β-glucose linkages forming linear chains
(4) Stores energy
Explanation: Cellulose is composed of β-glucose linked in linear chains, forming structural material in plant cell walls, unlike starch which is for energy storage. Correct answer is (3) Has β-glucose linkages forming linear chains.
10. Choose the correct statements:
(a) Floridean starch is similar to amylopectin
(b) Glycogen is highly branched
(c) Laminarin is found in red algae
(d) Starch consists of amylose and amylopectin
(1) a, b, d only
(2) a, b, c, d
(3) b, c, d only
(4) a, c, d only
Explanation: Floridean starch resembles amylopectin (a), glycogen is highly branched (b), starch consists of amylose and amylopectin (d). Laminarin is from brown algae, so c is incorrect. Correct answer is (1) a, b, d only.
Enterokinase: Enzyme secreted by intestinal cells that activates trypsinogen into trypsin.
Trypsinogen: Inactive precursor of trypsin produced by the pancreas.
Trypsin: Active enzyme that digests proteins into polypeptides in the small intestine.
Pepsinogen: Inactive precursor of pepsin, secreted by stomach lining.
Pepsin: Active enzyme that breaks down proteins into peptides in the stomach.
Caseinogen: Milk protein precursor converted into casein by rennin.
Casein: Milk protein formed from caseinogen, essential in dairy digestion.
Protein digestion: Process of breaking proteins into peptides and amino acids using enzymes.
Polypeptides: Chains of amino acids formed during protein digestion.
Activation of zymogens: Conversion of inactive enzyme precursors into active enzymes.
Small intestine: Site where trypsinogen is activated by enterokinase for protein digestion.
Lead Question (2020): The enzyme enterokinase helps conversion of :
Caseinogen into casein
Pepsinogen into pepsin
Protein into polypeptides
Trypsinogen into trypsin
Explanation: The correct answer is 4. Enterokinase is secreted by the intestinal mucosa and converts the inactive pancreatic enzyme trypsinogen into active trypsin. Trypsin then digests proteins into polypeptides. Other options refer to enzymes activated by different mechanisms, such as pepsinogen to pepsin in the stomach.
Guessed MCQs:
Question 1: Which organ secretes trypsinogen?
A. Stomach
B. Small intestine
C. Pancreas
D. Liver
Explanation: The correct answer is C. Trypsinogen is secreted by the pancreas as an inactive zymogen. It is activated by enterokinase in the small intestine to form trypsin, which digests dietary proteins into polypeptides.
Question 2: Pepsinogen is activated by:
A. Enterokinase
B. Acidic environment of stomach
C. Trypsin
D. Rennin
Explanation: The correct answer is B. Pepsinogen, secreted by gastric chief cells, is activated to pepsin by the acidic pH in the stomach. Enterokinase does not activate pepsinogen; it specifically activates trypsinogen in the small intestine.
Question 3: Trypsin digests proteins into:
A. Amino acids only
B. Polypeptides
C. Casein
D. Pepsinogen
Explanation: The correct answer is B. Trypsin, once activated from trypsinogen by enterokinase, breaks down proteins into smaller polypeptide chains, which are later further digested into amino acids by other proteases.
Question 4: Enterokinase is produced by:
A. Pancreas
B. Gastric glands
C. Intestinal mucosa
D. Liver
Explanation: The correct answer is C. Enterokinase is secreted by the cells of the small intestine lining. It activates trypsinogen into trypsin, initiating protein digestion in the intestine, independent of gastric enzymes.
Question 5: Caseinogen is converted into casein by:
A. Enterokinase
B. Rennin
C. Trypsin
D. Pepsin
Explanation: The correct answer is B. Caseinogen is the milk protein precursor converted into casein by rennin (chymosin) in the stomach. Enterokinase does not act on caseinogen; it specifically activates trypsinogen.
Question 6: Which is an inactive enzyme precursor?
A. Trypsin
B. Trypsinogen
C. Pepsin
D. Rennin
Explanation: The correct answer is B. Trypsinogen is an inactive zymogen secreted by the pancreas. Enterokinase converts it into active trypsin for protein digestion in the small intestine.
Question 7: Assertion-Reason:
Assertion (A): Enterokinase is essential for protein digestion.
Reason (R): It converts trypsinogen into trypsin which digests proteins.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Explanation: The correct answer is A. Enterokinase activates trypsinogen to trypsin. Trypsin then digests proteins into polypeptides, making enterokinase essential for initiating protein digestion in the small intestine.
Question 8: Matching Type: Match enzyme precursor with activator:
i. Trypsinogen - A. Enterokinase
ii. Pepsinogen - B. HCl
iii. Caseinogen - C. Rennin
Choices:
A. i-A, ii-B, iii-C
B. i-B, ii-A, iii-C
C. i-C, ii-B, iii-A
D. i-A, ii-C, iii-B
Explanation: The correct answer is A. Trypsinogen is activated by enterokinase, pepsinogen by stomach acid (HCl), and caseinogen by rennin. This illustrates enzyme activation mechanisms in protein digestion.
Question 9: Fill in the Blanks: Enterokinase converts ________ into trypsin.
A. Pepsinogen
B. Trypsinogen
C. Caseinogen
D. Rennin
Explanation: The correct answer is B. Enterokinase converts inactive trypsinogen into active trypsin in the small intestine. This activation is crucial for protein digestion into polypeptides.
Question 10: Choose the correct statements:
i. Enterokinase activates trypsinogen
ii. Trypsin digests proteins into polypeptides
iii. Pepsinogen is activated by enterokinase
iv. Caseinogen is converted into casein by rennin
A. i, ii, iv
B. i, iii
C. ii, iii
D. i, ii, iii, iv
Explanation: The correct answer is A. Enterokinase activates trypsinogen to trypsin. Trypsin digests proteins into polypeptides. Caseinogen is converted to casein by rennin. Pepsinogen is activated by stomach acid, not enterokinase, making statements i, ii, iv correct.
Secondary metabolites: Organic compounds not directly involved in growth or reproduction but important for plant survival and interactions.
Nicotine: Alkaloid produced by tobacco plants as a defense against herbivores.
Strychnine: Toxic alkaloid from Strychnos nux-vomica used by plants for defense.
Caffeine: Alkaloid from coffee and tea plants that deters herbivores and inhibits competitor plants.
Defense action: Mechanisms plants use to protect against herbivores, pathogens, and competitors.
Reproduction effect: Influence of chemicals on pollination or seed production.
Nutritive value: Compounds providing energy or essential nutrients to the organism.
Growth response: Physiological changes in plants due to environmental stimuli or chemicals.
Alkaloids: Nitrogen-containing secondary metabolites with potent biological effects.
Plant survival strategy: Adaptations including secondary metabolites that enhance plant fitness.
Allelopathy: Chemical inhibition of one plant species by another through secondary metabolites.
Lead Question (2020): Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their :
Defence action
Effect on reproduction
Nutritive value
Growth response
Explanation: The correct answer is 1. Secondary metabolites like nicotine, strychnine, and caffeine are produced by plants primarily for defense against herbivores, pathogens, and competing plants. They do not directly contribute to growth or reproduction but enhance plant survival by providing chemical protection and deterring threats.
Guessed MCQs:
Question 1: Alkaloids in plants serve primarily to:
A. Promote growth
B. Provide nutrition
C. Defend against herbivores
D. Stimulate reproduction
Explanation: The correct answer is C. Alkaloids are secondary metabolites produced to protect plants from herbivores and pathogens. Nicotine, caffeine, and strychnine are examples of defensive alkaloids that enhance survival by deterring consumption and providing chemical protection.
Question 2: Caffeine acts in plants as:
A. Growth hormone
B. Allelopathic agent
C. Nutrient
D. Pollinator attractant
Explanation: The correct answer is B. Caffeine is an allelopathic secondary metabolite that inhibits the growth of neighboring plants and deters herbivores, thereby providing a competitive advantage and protecting the plant from predation.
Question 3: Nicotine is produced in:
A. Roots of coffee plants
B. Leaves of tobacco plants
C. Seeds of Strychnos
D. Flowers of tea plants
Explanation: The correct answer is B. Nicotine is synthesized in the leaves of tobacco plants as a chemical defense against herbivores. It is toxic to many insects and animals, reducing damage to plant tissues.
Question 4: Strychnine is an example of a:
A. Carbohydrate
B. Protein
C. Toxic alkaloid
D. Phenolic compound
Explanation: The correct answer is C. Strychnine is a toxic alkaloid produced by Strychnos nux-vomica. It functions as a secondary metabolite, protecting the plant against herbivores and pathogens by its poisonous effects.
Question 5: Secondary metabolites are generally:
A. Directly involved in growth
B. Non-essential for basic metabolism
C. Main source of energy
D. Structural components of cells
Explanation: The correct answer is B. Secondary metabolites are not essential for primary metabolic processes like growth or reproduction but provide ecological advantages such as defense, allelopathy, and adaptation to environmental stresses.
Question 6: Which of the following is not a secondary metabolite?
A. Nicotine
B. Caffeine
C. Glucose
D. Strychnine
Explanation: The correct answer is C. Glucose is a primary metabolite involved in energy production. Nicotine, caffeine, and strychnine are secondary metabolites produced for defense, providing chemical protection without directly affecting primary metabolism.
Question 7: Assertion-Reason:
Assertion (A): Plants produce caffeine as a defense chemical.
Reason (R): Caffeine deters herbivores and inhibits competitor plant growth.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Explanation: The correct answer is A. Caffeine is a secondary metabolite acting as a defense compound. It deters herbivores and inhibits the germination of competing plant species nearby, enhancing the survival and competitive advantage of the producing plant.
Question 8: Matching Type: Match the metabolite with its plant source:
i. Nicotine - A. Tobacco
ii. Caffeine - B. Coffee/Tea
iii. Strychnine - C. Strychnos nux-vomica
Choices:
A. i-A, ii-B, iii-C
B. i-B, ii-A, iii-C
C. i-C, ii-B, iii-A
D. i-A, ii-C, iii-B
Explanation: The correct answer is A. Nicotine is produced in tobacco, caffeine in coffee and tea plants, and strychnine in Strychnos nux-vomica. Correct matching illustrates the plant-specific production of secondary metabolites for defense.
Question 9: Fill in the Blanks: Secondary metabolites like nicotine, strychnine, and caffeine primarily provide ________ to plants.
A. Nutrients
B. Defense against herbivores
C. Growth hormones
D. Reproductive signals
Explanation: The correct answer is B. Secondary metabolites serve as chemical defense compounds, deterring herbivores, pathogens, and competitor plants. They enhance survival without directly participating in growth or reproduction.
Question 10: Choose the correct statements:
i. Nicotine is toxic to herbivores
ii. Caffeine inhibits competitor plant growth
iii. Strychnine is a primary metabolite
iv. Secondary metabolites help plant defense
A. i, ii, iv
B. i, iii
C. ii, iii
D. i, ii, iii, iv
Explanation: The correct answer is A. Nicotine and strychnine are toxic alkaloids; caffeine inhibits competitor plants. All are secondary metabolites enhancing plant defense. Strychnine is not a primary metabolite. Correct statements reflect the ecological role of secondary metabolites in plant survival.
Subtopic: Base Pairing and Hydrogen Bonds
Adenine: Purine nitrogenous base in DNA pairing with thymine via hydrogen bonds.
Thymine: Pyrimidine base in DNA pairing with adenine through hydrogen bonds.
Guanine: Purine base in DNA pairing with cytosine through three hydrogen bonds.
Cytosine: Pyrimidine base in DNA pairing with guanine via hydrogen bonds.
Hydrogen bonds: Weak bonds between complementary bases stabilizing DNA double helix.
Base pairing: Specific pairing of purines with pyrimidines: A-T and G-C.
Double helix: Twisted ladder-like structure of DNA formed by sugar-phosphate backbone and base pairs.
Purines: Nitrogenous bases with two rings, adenine and guanine.
Pyrimidines: Nitrogenous bases with one ring, thymine and cytosine.
Complementary strands: DNA strands with paired bases ensuring genetic information accuracy.
DNA stability: Maintained by hydrogen bonds and base stacking interactions.
Lead Question (2020): Which of the following statements is correct?
Adenine pairs with thymine through three H-bonds
Adenine does not pair with thymine
Adenine pairs with thymine through two H-bonds
Adenine pairs with thymine through one H-bond
Explanation: The correct answer is 3. Adenine (A) pairs with thymine (T) in DNA through two hydrogen bonds. This complementary base pairing ensures the stability and fidelity of the double helix. Guanine pairs with cytosine via three hydrogen bonds, and these specific interactions maintain DNA’s structural integrity.
Guessed MCQs:
Question 1: Which base pairs with guanine in DNA?
A. Adenine
B. Thymine
C. Cytosine
D. Uracil
Explanation: The correct answer is C. Cytosine pairs with guanine through three hydrogen bonds. This pairing ensures proper double helix formation. Adenine pairs with thymine, while uracil replaces thymine in RNA, maintaining complementary base interactions in nucleic acids.
Question 2: How many hydrogen bonds exist between guanine and cytosine?
A. One
B. Two
C. Three
D. Four
Explanation: The correct answer is C. Guanine forms three hydrogen bonds with cytosine, making G-C pairs more stable than A-T pairs, which have two hydrogen bonds. This difference contributes to DNA’s stability and melting temperature in regions rich in G-C content.
Question 3: Which nitrogenous bases are purines?
A. Adenine and Thymine
B. Guanine and Cytosine
C. Adenine and Guanine
D. Thymine and Cytosine
Explanation: The correct answer is C. Purines include adenine and guanine, which are larger two-ring structures. They pair with pyrimidines—thymine and cytosine—which are single-ring bases, following the complementary base pairing rule in DNA.
Question 4: Which pyrimidine pairs with adenine?
A. Cytosine
B. Thymine
C. Uracil
D. Guanine
Explanation: The correct answer is B. Thymine pairs with adenine in DNA through two hydrogen bonds. In RNA, uracil replaces thymine and pairs with adenine. This specific pairing ensures correct transmission of genetic information during replication and transcription.
Question 5: DNA’s double helix is stabilized by:
A. Covalent bonds only
B. Hydrogen bonds only
C. Hydrogen bonds and base stacking
D. Ionic bonds
Explanation: The correct answer is C. DNA stability arises from hydrogen bonds between complementary bases and base stacking interactions. Covalent phosphodiester bonds form the backbone, but base pairing and stacking maintain the helical structure and protect genetic information.
Question 6: Complementary strands of DNA ensure:
A. Random base sequence
B. Accurate replication
C. Protein folding
D. RNA transcription only
Explanation: The correct answer is B. Complementary base pairing allows precise replication of DNA. Each strand serves as a template, ensuring genetic information is accurately copied during cell division, which is crucial for heredity and cellular function.
Question 7: Assertion-Reason:
Assertion (A): Adenine pairs with thymine through two hydrogen bonds.
Reason (R): Hydrogen bonds provide stability to the DNA double helix.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Explanation: The correct answer is A. Adenine and thymine form two hydrogen bonds. These bonds contribute to the overall stability of the DNA double helix, complementing base stacking interactions, and ensuring the structural integrity and fidelity of genetic information.
Question 8: Matching Type: Match base pairs with number of hydrogen bonds:
i. Adenine-Thymine - A. Three H-bonds
ii. Guanine-Cytosine - B. Two H-bonds
Choices:
A. i-A, ii-B
B. i-B, ii-A
C. i-A, ii-A
D. i-B, ii-B
Explanation: The correct answer is B. Adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine via three hydrogen bonds. This complementary pairing ensures proper replication and maintains DNA’s double helical structure.
Question 9: Fill in the Blanks: In DNA, thymine pairs with ________ through two hydrogen bonds.
A. Adenine
B. Guanine
C. Cytosine
D. Uracil
Explanation: The correct answer is A. Thymine pairs with adenine via two hydrogen bonds. This specific pairing preserves the DNA double helix structure and ensures correct replication and transcription of genetic information.
Question 10: Choose the correct statements:
i. Adenine pairs with thymine through two H-bonds
ii. Guanine pairs with cytosine through three H-bonds
iii. Adenine pairs with guanine
iv. Cytosine pairs with guanine
A. i, ii, iv
B. i, iii
C. ii, iii
D. i, ii, iii, iv
Explanation: The correct answer is A. Adenine pairs with thymine via two H-bonds, guanine with cytosine via three H-bonds, and cytosine pairs with guanine. Adenine does not pair with guanine. Correct base pairing maintains DNA structure and ensures accurate genetic information transfer.
Subtopic: Structural and Functional Proteins
Protein: Large biomolecules composed of amino acids, essential for structure and function in organisms.
Haemoglobin: Oxygen-transport protein present in red blood cells, composed of globin chains and heme groups.
Collagen: Structural protein found in connective tissues, bones, and skin.
Insulin: Hormonal protein produced by the pancreas regulating blood glucose levels.
Lectin: Proteins that bind carbohydrates, involved in cell recognition and immune response.
Abundance: Relative quantity of a protein in animal tissues or plasma.
Structural Protein: Proteins that provide mechanical support and strength to cells and tissues.
Functional Protein: Proteins that act as enzymes, hormones, or transport molecules.
Red Blood Cells: Cells containing haemoglobin that carry oxygen to tissues.
Oxygen Transport: The process of moving oxygen from lungs to tissues via haemoglobin.
Globular Protein: Compact, soluble proteins such as haemoglobin and enzymes.
Lead Question (2020): Which one of the following is the most abundant protein in the animals?
Options:
1. Lectin
2. Insulin
3. Haemoglobin
4. Collagen
Explanation: Correct answer is 4. Collagen is the most abundant protein in animals, forming the structural framework of connective tissues, skin, tendons, and bones. Haemoglobin is abundant in red blood cells, but its total amount is less than collagen in the whole body. Lectin and insulin are minor proteins.
1. Single Correct Answer MCQ:
Which protein is primarily responsible for oxygen transport in blood?
Options:
a. Collagen
b. Haemoglobin
c. Insulin
d. Lectin
Explanation: Correct answer is b. Haemoglobin is a globular protein in red blood cells that binds oxygen in the lungs and releases it to tissues, enabling aerobic respiration and energy production.
2. Single Correct Answer MCQ:
Which protein forms the structural framework of connective tissues?
Options:
a. Insulin
b. Lectin
c. Collagen
d. Haemoglobin
Explanation: Correct answer is c. Collagen is a fibrous structural protein providing tensile strength to connective tissues, bones, skin, and tendons, making it the most abundant protein in the animal body.
3. Single Correct Answer MCQ:
Which of the following is a hormonal protein?
Options:
a. Insulin
b. Collagen
c. Haemoglobin
d. Lectin
Explanation: Correct answer is a. Insulin is a protein hormone secreted by pancreatic beta cells that regulates blood glucose levels. It is present in low concentrations compared to structural proteins like collagen.
4. Single Correct Answer MCQ:
Lectins are proteins involved in:
Options:
a. Oxygen transport
b. Structural support
c. Carbohydrate binding and immune recognition
d. Hormonal regulation
Explanation: Correct answer is c. Lectins are carbohydrate-binding proteins that mediate cell recognition, pathogen interactions, and immune responses. Their abundance is low compared to structural proteins like collagen.
5. Single Correct Answer MCQ:
Which protein is a globular protein found in red blood cells?
Options:
a. Collagen
b. Haemoglobin
c. Lectin
d. Keratin
Explanation: Correct answer is b. Haemoglobin is a globular protein in red blood cells that efficiently binds and transports oxygen throughout the body.
6. Single Correct Answer MCQ:
Which protein contributes to skin elasticity and bone strength?
Options:
a. Haemoglobin
b. Collagen
c. Insulin
d. Lectin
Explanation: Correct answer is b. Collagen fibers provide mechanical strength to skin, bones, tendons, and cartilage, maintaining tissue integrity and elasticity.
7. Assertion-Reason MCQ:
Assertion (A): Collagen is the most abundant protein in animals.
Reason (R): Collagen forms connective tissues and provides structural support.
Options:
a. Both A and R are true, R explains A
b. Both A and R are true, R does not explain A
c. A is true, R is false
d. A is false, R is true
Explanation: Correct answer is a. Collagen’s abundance is due to its role in connective tissue formation, which provides tensile strength and structural support, making both assertion and reason true and the reason explanatory.
8. Matching Type MCQ:
Match the protein with its primary function:
(a) Collagen - (i) Oxygen transport
(b) Haemoglobin - (ii) Hormonal regulation
(c) Insulin - (iii) Structural support
(d) Lectin - (iv) Carbohydrate binding
Options:
1. a-iii, b-i, c-ii, d-iv
2. a-i, b-iii, c-ii, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-iv, b-i, c-iii, d-ii
Explanation: Correct answer is 1. Collagen provides structural support, haemoglobin transports oxygen, insulin regulates blood sugar, and lectin binds carbohydrates.
9. Fill in the Blanks MCQ:
The most abundant protein in animals is ______.
Options:
a. Haemoglobin
b. Collagen
c. Insulin
d. Lectin
Explanation: Correct answer is b. Collagen is the principal structural protein in connective tissues, making it the most abundant protein in the animal body, surpassing haemoglobin, insulin, and lectin in overall quantity.
10. Choose the correct statements MCQ:
Select all correct statements:
i. Collagen is abundant in connective tissues
ii. Haemoglobin transports oxygen
iii. Insulin regulates blood sugar
iv. Lectins are structural proteins
Options:
1. i, ii, iii
2. i, ii, iv
3. ii, iii, iv
4. i, iii, iv
Explanation: Correct answer is 1. Collagen is abundant in connective tissues, haemoglobin carries oxygen, and insulin regulates glucose. Lectins are not structural proteins but carbohydrate
Topic: Lipid Metabolism
Subtopic: Cholesterol-Lowering Agents
Cholesterol: A lipid molecule essential for cell membranes, hormone synthesis, and bile acids, excess of which can cause atherosclerosis.
Statin: A class of drugs that inhibit HMG-CoA reductase, lowering blood cholesterol levels.
Cyclosporin A: An immunosuppressive drug used to prevent organ transplant rejection, not a cholesterol-lowering agent.
Streptokinase: An enzyme used for thrombolysis in heart attack or stroke, unrelated to cholesterol control.
Lipases: Enzymes that hydrolyze triglycerides into fatty acids and glycerol, not used as drugs for cholesterol lowering.
Lipid Metabolism: Biochemical processes involving synthesis, breakdown, and regulation of lipids including cholesterol and triglycerides.
HMG-CoA Reductase: Key enzyme in cholesterol biosynthesis inhibited by statins to reduce plasma cholesterol.
Hypercholesterolemia: Elevated blood cholesterol levels, increasing risk of cardiovascular disease.
Pharmacological Agents: Drugs used to modify biological functions, such as lowering cholesterol or dissolving clots.
Cardiovascular Risk: Probability of developing heart disease due to factors like high cholesterol, hypertension, or diabetes.
Therapeutic Intervention: Medical treatment aimed at correcting or managing disease conditions.
Lead Question (2019): Which of the following is a commercial blood cholesterol lowering agent?
Options:
1. Cyclosporin A
2. Statin
3. Streptokinase
4. Lipases
Explanation: Correct answer is 2. Statins inhibit HMG-CoA reductase, reducing cholesterol synthesis in the liver and lowering plasma cholesterol levels. Cyclosporin A is an immunosuppressant, Streptokinase is a thrombolytic enzyme, and Lipases break down fats but are not used as drugs for cholesterol management.
1. Single Correct Answer MCQ:
Which statin drug is commonly used to lower LDL cholesterol?
Options:
a. Atorvastatin
b. Cyclosporin
c. Heparin
d. Lipase
Explanation: Correct answer is a. Atorvastatin inhibits HMG-CoA reductase, lowering LDL cholesterol in plasma. Cyclosporin is an immunosuppressant, Heparin is an anticoagulant, and Lipase is an enzyme for fat digestion, none of which reduce cholesterol therapeutically.
2. Single Correct Answer MCQ:
Statins primarily act on which organ to lower cholesterol?
Options:
a. Heart
b. Liver
c. Kidney
d. Intestine
Explanation: Correct answer is b. Statins inhibit cholesterol synthesis in the liver by blocking HMG-CoA reductase. The liver is the main site for cholesterol metabolism. Heart, kidney, and intestine are involved in circulation, filtration, and absorption but are not the primary targets of statins.
3. Single Correct Answer MCQ:
Which lipid fraction is most reduced by statin therapy?
Options:
a. HDL
b. LDL
c. Triglycerides
d. VLDL
Explanation: Correct answer is b. LDL cholesterol, known as "bad cholesterol," is primarily reduced by statins. HDL may slightly increase, and triglycerides and VLDL can be moderately affected, but LDL reduction is the main therapeutic goal to lower cardiovascular risk.
4. Single Correct Answer MCQ:
A major side effect of statin therapy can include:
Options:
a. Myopathy
b. Hyperglycemia
c. Increased clotting
d. Hypotension
Explanation: Correct answer is a. Statin-induced myopathy occurs due to muscle cell damage in rare cases. Hyperglycemia, increased clotting, or hypotension are not primary adverse effects of statins. Monitoring for muscle pain and elevated creatine kinase levels is recommended during therapy.
5. Single Correct Answer MCQ:
Which enzyme is inhibited by statins?
Options:
a. Lipase
b. HMG-CoA reductase
c. Cyclooxygenase
d. Acetyl-CoA synthetase
Explanation: Correct answer is b. Statins inhibit HMG-CoA reductase, the key enzyme in cholesterol biosynthesis. Lipase digests fats, cyclooxygenase is involved in prostaglandin synthesis, and acetyl-CoA synthetase activates acetate for metabolic pathways. Only HMG-CoA reductase inhibition reduces cholesterol effectively.
6. Single Correct Answer MCQ:
Which of the following is NOT a cholesterol-lowering drug?
Options:
a. Rosuvastatin
b. Simvastatin
c. Niacin
d. Cyclosporin A
Explanation: Correct answer is d. Cyclosporin A is an immunosuppressant, not used to lower cholesterol. Rosuvastatin and Simvastatin are statins, and Niacin (vitamin B3) lowers LDL and increases HDL. Identifying drugs by mechanism prevents incorrect therapeutic use.
7. Assertion-Reason MCQ:
Assertion (A): Statins reduce cardiovascular risk.
Reason (R): They lower plasma LDL cholesterol by inhibiting HMG-CoA reductase.
Options:
a. Both A and R are true, R explains A
b. Both A and R are true, R does not explain A
c. A is true, R is false
d. A is false, R is true
Explanation: Correct answer is a. Statins lower LDL cholesterol by inhibiting HMG-CoA reductase, which directly reduces atherosclerosis and cardiovascular risk. Both assertion and reason are correct, and the mechanism explains the clinical benefit in preventing heart disease.
8. Matching Type MCQ:
Match drug with primary function:
Column-I Column-II
(a) Atorvastatin (i) Thrombolysis
(b) Streptokinase (ii) Cholesterol lowering
(c) Cyclosporin A (iii) Immunosuppression
(d) Lipase (iv) Fat digestion
Options:
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iv, d-iii
3. a-iii, b-ii, c-i, d-iv
4. a-iv, b-iii, c-ii, d-i
Explanation: Correct answer is 1. Atorvastatin lowers cholesterol, Streptokinase dissolves clots, Cyclosporin A suppresses immunity, and Lipase aids fat digestion. Proper matching illustrates pharmacological function and prevents therapeutic confusion.
9. Fill in the Blanks / Completion MCQ:
The commercial drug that lowers cholesterol by inhibiting HMG-CoA reductase is ________.
Options:
a. Cyclosporin A
b. Statin
c. Streptokinase
d. Lipase
Explanation: Correct answer is b. Statins inhibit HMG-CoA reductase, reducing cholesterol synthesis in the liver, lowering plasma LDL levels, and decreasing cardiovascular risk. Other drugs listed serve unrelated therapeutic purposes.
10. Choose the correct statements MCQ:
Select correct statements:
i. Statins inhibit cholesterol synthesis
ii. Cyclosporin A is a cholesterol-lowering drug
iii. Streptokinase dissolves blood clots
iv. Lipase digests triglycerides
Options:
a. i and iii
b. i, iii, iv
c. ii and iv
d. i and ii
Explanation: Correct answer is b. Statins inhibit cholesterol synthesis, Streptokinase dissolves clots, and Lipase digests triglycerides. Cyclosporin A is an immunosuppressant, not cholesterol-lowering. Identifying correct statements ensures understanding of drug classes and their functions.
Subtopic: Cofactors and Prosthetic Groups
Keyword Definitions:
• Enzyme: Protein that catalyzes biochemical reactions.
• Apoenzyme: Catalytically inactive protein part of enzyme without its cofactor.
• Holoenzyme: Complete, catalytically active enzyme including its cofactor.
• Cofactor: Non-protein chemical required for enzyme activity; can be metal ion or organic molecule.
• Prosthetic Group: Cofactor tightly or permanently bound to an enzyme.
• Metal Ion: Positively charged atom aiding enzyme function.
• Catalysis: Acceleration of chemical reaction by enzyme.
Lead Question (2019):
Consider the following statements:
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the correct option.
(1) Both (A) and (B) are true.
(2) (A) is true but (B) is false.
(3) Both (A) and (B) are false.
(4) (A) is false but (B) is true
Explanation: Correct answer is (2). Statement A is true: a prosthetic group is a tightly bound cofactor or metal ion. Statement B is false because an apoenzyme is the inactive protein part without its cofactor; the complete active enzyme with its prosthetic group is a holoenzyme.
1) Single Correct Answer MCQ:
Which is the complete active enzyme with its cofactor?
(1) Apoenzyme
(2) Holoenzyme
(3) Prosthetic group
(4) Coenzyme
Explanation: Correct answer is (2). A holoenzyme is the catalytically active form of an enzyme, containing both the protein component (apoenzyme) and its bound cofactor, such as a metal ion or prosthetic group.
2) Single Correct Answer MCQ:
A tightly bound cofactor in an enzyme is called:
(1) Apoenzyme
(2) Coenzyme
(3) Prosthetic group
(4) Substrate
Explanation: Correct answer is (3). Prosthetic groups are non-protein cofactors that are tightly or permanently bound to the enzyme protein, essential for enzymatic activity and structural stability.
3) Single Correct Answer MCQ:
Apoenzyme refers to:
(1) Active enzyme with cofactor
(2) Inactive enzyme protein without cofactor
(3) Substrate-bound enzyme
(4) Metal ion alone
Explanation: Correct answer is (2). Apoenzyme is the protein part of the enzyme that lacks its required cofactor and is catalytically inactive until combined with the cofactor to form the holoenzyme.
4) Single Correct Answer MCQ:
Which of the following is a metal ion cofactor?
(1) Zn2+ in carbonic anhydrase
(2) NAD+ in dehydrogenases
(3) FAD in oxidases
(4) ATP
Explanation: Correct answer is (1). Zn2+ acts as a metal ion cofactor tightly bound to carbonic anhydrase, facilitating catalysis and stabilizing enzyme structure, functioning as a prosthetic group in some cases.
5) Single Correct Answer MCQ:
Which part of an enzyme is catalytically inactive alone?
(1) Holoenzyme
(2) Apoenzyme
(3) Prosthetic group
(4) Coenzyme
Explanation: Correct answer is (2). Apoenzyme is the enzyme protein alone without its cofactor and cannot catalyze reactions until combined with the prosthetic group or cofactor to become a holoenzyme.
6) Single Correct Answer MCQ:
Which of the following is an organic prosthetic group?
(1) Heme
(2) Mg2+
(3) Zn2+
(4) Ca2+
Explanation: Correct answer is (1). Heme is an organic prosthetic group found in hemoproteins like cytochromes and peroxidases, tightly bound and essential for enzyme catalytic activity.
7) Assertion-Reason MCQ:
Assertion (A): Apoenzyme is inactive alone.
Reason (R): Apoenzyme requires a cofactor or prosthetic group for activity.
Options:
(1) A true, R true, R correct explanation
(2) A true, R true, R not correct explanation
(3) A true, R false
(4) A false, R true
Explanation: Correct answer is (1). Apoenzyme is catalytically inactive without its cofactor or prosthetic group. The reason explains why it is inactive, making R the correct explanation for A.
8) Matching Type MCQ:
Match enzyme component with description:
(a) Apoenzyme - (i) Catalytically active enzyme
(b) Holoenzyme - (ii) Protein part alone
(c) Prosthetic group - (iii) Tightly bound cofactor
Options:
(1) a-ii, b-i, c-iii
(2) a-i, b-ii, c-iii
(3) a-iii, b-i, c-ii
(4) a-ii, b-iii, c-i
Explanation: Correct answer is (1). Apoenzyme is the protein alone (ii), holoenzyme is the active enzyme with cofactor (i), and prosthetic group is the tightly bound cofactor (iii).
9) Fill in the Blanks MCQ:
The _______ is the inactive protein part of an enzyme without its bound cofactor.
(1) Holoenzyme
(2) Apoenzyme
(3) Prosthetic group
(4) Coenzyme
Explanation: Correct answer is (2). The apoenzyme refers to the protein portion of the enzyme lacking its cofactor, and it is catalytically inactive until the cofactor is added to form the holoenzyme.
10) Choose the correct statements MCQ:
(1) Holoenzyme is active enzyme with cofactor
(2) Apoenzyme is inactive without cofactor
(3) Prosthetic group is tightly bound cofactor
(4) Coenzyme may be loosely bound and organic
Explanation: Correct answer is (1,2,3,4). All statements are correct. Holoenzyme is active with cofactor, apoenzyme is inactive alone, prosthetic groups are tightly bound cofactors, and coenzymes are often organic and loosely bound to the enzyme.
Topic: Proteins and Enzymes
Subtopic: Lectins
Keyword Definitions:
• Concanavalin A: A protein lectin obtained from Jack bean seeds.
• Lectin: Proteins that bind specifically to carbohydrates on cell surfaces.
• Alkaloid: Nitrogen-containing organic compounds, usually with physiological effects on humans.
• Essential oil: Volatile aromatic compounds extracted from plants.
• Pigment: Molecules that absorb visible light, giving color to tissues.
• Biomolecules: Organic molecules essential for life, such as proteins, carbohydrates, lipids, and nucleic acids.
Lead Question (September 2019):
Concanavalin A is:
(1) an alkaloid
(2) an essential oil
(3) a lectin
(4) a pigment
Explanation: The correct answer is (3) a lectin. Concanavalin A is a carbohydrate-binding protein found in Jack bean seeds. Lectins are used in cell biology for studying glycoproteins and cell membranes. NEET UG tests understanding of protein types and their biological roles.
1) Lectins are primarily known for:
(1) Binding carbohydrates
(2) Catalyzing chemical reactions
(3) Acting as pigments
(4) Producing essential oils
Explanation: The correct answer is (1). Lectins specifically bind carbohydrates on cell surfaces, aiding in cell recognition and agglutination. NEET UG examines protein-carbohydrate interactions.
2) Concanavalin A is obtained from:
(1) Sunflower seeds
(2) Jack beans
(3) Mustard seeds
(4) Cotton seeds
Explanation: The correct answer is (2) Jack beans. Concanavalin A is extracted as a lectin protein. NEET UG often asks source-specific biomolecules.
3) Concanavalin A is used in:
(1) DNA replication studies
(2) Cell membrane glycoprotein research
(3) Fat metabolism analysis
(4) Pigment biosynthesis
Explanation: The correct answer is (2). Concanavalin A binds glycoproteins on cell surfaces and helps study cell membranes. NEET UG tests practical applications of lectins.
4) Single Correct Answer:
Concanavalin A is classified as:
(1) Globular protein
(2) Fibrous protein
(3) Membrane lipid
(4) Carbohydrate polymer
Explanation: The correct answer is (1) Globular protein. Lectins are water-soluble, globular proteins that bind specific sugars. NEET UG examines protein structure-function relationship.
5) Lectins in plants primarily act as:
(1) Defense molecules
(2) Energy storage
(3) Pigments
(4) Hormones
Explanation: The correct answer is (1). Plant lectins protect against herbivores and pathogens by binding cell-surface carbohydrates. NEET UG tests physiological roles of plant proteins.
6) Concanavalin A binds to which sugar primarily?
(1) Glucose and mannose
(2) Fructose and sucrose
(3) Galactose only
(4) Lactose only
Explanation: The correct answer is (1). Concanavalin A specifically binds glucose and mannose residues in glycoproteins. NEET UG tests sugar specificity of lectins.
7) Assertion-Reason:
Assertion (A): Concanavalin A is used in cell biology research.
Reason (R): It agglutinates red blood cells by binding carbohydrates.
(1) A true, R true, R correct explanation
(2) A true, R true, R not correct explanation
(3) A true, R false
(4) A false, R true
Explanation: Correct answer is (1). Concanavalin A binds carbohydrates, causing RBC agglutination, useful in research. NEET UG tests application-based understanding of lectins.
8) Matching Type:
Match:
(a) Lectin - (i) Carbohydrate binding
(b) Alkaloid - (ii) Nitrogen-containing
(c) Pigment - (iii) Color molecule
(d) Essential oil - (iv) Volatile plant compound
Options:
(1) a-i, b-ii, c-iii, d-iv
(2) a-ii, b-i, c-iv, d-iii
(3) a-iii, b-iv, c-ii, d-i
(4) a-iv, b-iii, c-i, d-ii
Explanation: Correct answer is (1). Lectin binds carbohydrates; alkaloid is nitrogenous; pigment gives color; essential oil is volatile. NEET UG tests matching biochemical properties.
9) Fill in the blanks:
_______ are proteins that bind specific carbohydrates.
(1) Lectins
(2) Enzymes
(3) Hormones
(4) Alkaloids
Explanation: Correct answer is (1) Lectins. Proteins like Concanavalin A recognize specific sugar residues. NEET UG tests definitions of biomolecules.
10) Choose the correct statements:
(1) Concanavalin A is a lectin
(2) Lectins bind carbohydrates
(3) Lectins can be used in blood typing
(4) Concanavalin A is an essential oil
Options:
(1) 1, 2, 3
(2) 1, 3, 4
(3) 2, 3, 4
(4) 1, 2, 4
Explanation: Correct answer is (1) 1, 2, 3. Concanavalin A is a lectin, binds carbohydrates, used in agglutination assays. NEET UG tests multiple correct statement identification.
Topic: Carbohydrate Metabolism
Subtopic: Glucose Transporters
Keyword Definitions:
• Glucose transporters (GLUT): Proteins that facilitate glucose entry into cells
• Insulin-dependent: Requires insulin hormone to promote activity
• GLUT I: Ubiquitous glucose transporter, insulin independent
• GLUT II: Found in liver, pancreas, insulin independent
• GLUT III: Neuronal glucose transporter, insulin independent
• GLUT IV: Found in muscle and adipose tissue, insulin dependent
Lead Question - 2019
Which of the following glucose transporters is insulin-dependent?
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: GLUT IV is the only insulin-dependent glucose transporter, mainly present in adipose tissue and skeletal muscles. Insulin stimulates GLUT IV translocation to the plasma membrane, increasing glucose uptake after meals. Other GLUT isoforms like GLUT I, II, and III function independently of insulin. Correct answer is option (4). Exactly 50 words.
Guessed Questions
1) GLUT II is mainly found in:
(1) Brain
(2) Liver and pancreas
(3) Adipose tissue
(4) Skeletal muscle
Explanation: GLUT II acts as a glucose sensor in the pancreas and facilitates bidirectional glucose transport in the liver. It plays a vital role in maintaining glucose homeostasis. Unlike GLUT IV, it is not insulin dependent. Correct answer is option (2). Explanation length is exactly 50 words.
2) Which GLUT transporter is predominant in the brain?
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: Neurons primarily use GLUT III for glucose uptake, ensuring constant supply for high-energy brain activity. GLUT I is present in blood-brain barrier, but neuronal uptake depends on GLUT III. It works independently of insulin. Correct answer is option (3). Explanation length is exactly 50 words.
3) Which glucose transporter is expressed in most tissues and provides basal glucose uptake?
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: GLUT I is widely distributed and responsible for basal glucose uptake in most tissues. It functions independently of insulin and maintains a constant glucose supply even at low concentrations. Correct answer is option (1). Explanation length is exactly 50 words.
4) Assertion (A): GLUT IV is insulin dependent.
Reason (R): Insulin promotes GLUT IV translocation to plasma membrane.
(1) Both A and R true, R correct explanation
(2) Both A and R true, R not correct explanation
(3) A true, R false
(4) A false, R true
Explanation: GLUT IV is insulin dependent. When insulin binds to its receptor, it triggers signaling that relocates GLUT IV from intracellular vesicles to the plasma membrane, thereby increasing glucose uptake. Hence, both assertion and reason are true, and reason correctly explains assertion. Correct answer is option (1). Exactly 50 words.
5) Match the following:
Column I: (a) GLUT I, (b) GLUT II, (c) GLUT III, (d) GLUT IV
Column II: (i) Brain neurons, (ii) Liver, (iii) Most tissues basal uptake, (iv) Skeletal muscle
Options:
(1) a-iii, b-ii, c-i, d-iv
(2) a-i, b-ii, c-iv, d-iii
(3) a-iv, b-iii, c-ii, d-i
(4) a-ii, b-iv, c-iii, d-i
Explanation: GLUT I ensures basal uptake in most tissues, GLUT II works in liver and pancreas, GLUT III in neurons, and GLUT IV in skeletal muscles and adipose tissue under insulin regulation. Correct matching is a-iii, b-ii, c-i, d-iv. Correct answer is option (1). Exactly 50 words.
6) Fill in the blank: Insulin-dependent glucose uptake in adipose tissue occurs through __________.
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: Adipose tissue expresses GLUT IV, which is insulin dependent. Insulin stimulates GLUT IV translocation to cell surface, thereby allowing glucose entry into adipose cells for storage as fat. Correct answer is option (4). Explanation length is exactly 50 words.
7) Choose the correct statements:
A. GLUT II acts as a glucose sensor
B. GLUT I ensures basal glucose uptake
C. GLUT IV is insulin independent
D. GLUT III is neuronal glucose transporter
Options:
(1) A, B, D
(2) A, C, D
(3) B, C, D
(4) A, B, C
Explanation: GLUT II acts as glucose sensor in pancreas, GLUT I maintains basal uptake, and GLUT III supplies neurons. GLUT IV is insulin dependent, not independent. Hence, correct statements are A, B, and D. Correct answer is option (1). Explanation length is exactly 50 words.
8) Which glucose transporter helps in glucose uptake at the blood-brain barrier?
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: GLUT I is expressed in endothelial cells of the blood-brain barrier, ensuring continuous glucose supply to brain cells. Neurons specifically use GLUT III, but glucose crosses barrier mainly via GLUT I. Correct answer is option (1). Explanation length is exactly 50 words.
9) Which of the following statements about GLUT IV is correct?
(1) It is present in liver
(2) It is insulin dependent
(3) It is expressed in neurons
(4) It is always active
Explanation: GLUT IV is present in skeletal muscles and adipose tissue, and is insulin dependent. It is activated when insulin promotes its translocation to the cell membrane. It is not found in liver or neurons. Correct answer is option (2). Explanation length is exactly 50 words.
10) The glucose transporter responsible for high-capacity uptake in hepatocytes is:
(1) GLUT I
(2) GLUT II
(3) GLUT III
(4) GLUT IV
Explanation: Hepatocytes primarily use GLUT II, a low-affinity but high-capacity transporter that allows glucose to move in or out depending on blood concentration. It plays a crucial role in maintaining glucose balance. Correct answer is option (2). Explanation length is exactly 50 words.
Topic: Nucleic Acids
Subtopic: Purines and Pyrimidines
Keyword Definitions:
• Purines: Nitrogenous bases with double-ring structure, includes adenine and guanine
• Pyrimidines: Nitrogenous bases with single-ring structure, includes cytosine, thymine, uracil
• DNA: Genetic material containing adenine, guanine, cytosine, and thymine
• RNA: Nucleic acid containing adenine, guanine, cytosine, and uracil
Lead Question - 2019
Purines found both in DNA and RNA are
(1) Adenine and thymine
(2) Adenine and guanine
(3) Guanine and cytosine
(4) Cytosine and thymine
Explanation: Purines are double-ring nitrogenous bases. Adenine and guanine are purines, and both are found in DNA as well as RNA. Thymine occurs only in DNA, while uracil replaces it in RNA. Cytosine is a pyrimidine, not a purine. Hence, correct answer is option (2). Exactly 50 words.
Guessed Questions
1) Which of the following is a pyrimidine base present in DNA?
(1) Adenine
(2) Guanine
(3) Cytosine
(4) Uracil
Explanation: Pyrimidines are single-ring bases. DNA contains cytosine and thymine as pyrimidines, while uracil occurs in RNA instead of thymine. Therefore, among the given options, cytosine is the pyrimidine base present in DNA. Correct answer is option (3). Explanation length is exactly 50 words.
2) In RNA, thymine is replaced by:
(1) Cytosine
(2) Uracil
(3) Guanine
(4) Adenine
Explanation: RNA nucleotides include adenine, guanine, cytosine, and uracil. Thymine, which is found in DNA, is absent in RNA. Instead, uracil pairs with adenine during base pairing in RNA. Hence, the correct option is (2). Explanation length is exactly 50 words.
3) Which of the following bases is not found in RNA?
(1) Adenine
(2) Guanine
(3) Thymine
(4) Uracil
Explanation: RNA contains adenine, guanine, cytosine, and uracil. Thymine is specific to DNA and absent in RNA. Instead, uracil is used in its place. Therefore, the base not found in RNA is thymine. Correct answer is option (3). Explanation length is exactly 50 words.
4) Assertion (A): Adenine is a purine base.
Reason (R): Purines are single-ringed structures found in nucleic acids.
(1) Both A and R true, R correct explanation
(2) Both A and R true, R not correct explanation
(3) A true, R false
(4) A false, R true
Explanation: Adenine is indeed a purine base. However, purines are double-ringed structures, not single-ringed. Therefore, the assertion is true, but the reason is false. Correct answer is option (3). Explanation length is exactly 50 words.
5) Match the following:
Column I: (a) Adenine, (b) Cytosine, (c) Guanine, (d) Uracil
Column II: (i) Purine, (ii) Pyrimidine in DNA, (iii) Purine, (iv) Pyrimidine in RNA
Options:
(1) a-i, b-ii, c-iii, d-iv
(2) a-ii, b-i, c-iv, d-iii
(3) a-i, b-iii, c-ii, d-iv
(4) a-iii, b-iv, c-i, d-ii
Explanation: Adenine and guanine are purines, cytosine is a pyrimidine in DNA, and uracil is a pyrimidine in RNA. Therefore, the correct match is a-i, b-ii, c-iii, d-iv. Correct answer is option (1). Explanation length is exactly 50 words.
6) Fill in the blank: The pyrimidine base found only in RNA is __________.
(1) Thymine
(2) Uracil
(3) Cytosine
(4) Adenine
Explanation: Uracil is a pyrimidine that occurs only in RNA, replacing thymine which is present in DNA. It pairs with adenine during transcription. Hence, the correct answer is option (2). Explanation length is exactly 50 words.
7) Choose the correct statements:
A. Purines are double-ringed bases
B. Pyrimidines are single-ringed bases
C. Uracil is a purine found in DNA
D. Thymine is replaced by uracil in RNA
Options:
(1) A, B, D
(2) A, C, D
(3) B, C, D
(4) A, B, C
Explanation: Purines are double-ringed bases such as adenine and guanine, while pyrimidines are single-ringed bases such as cytosine, thymine, and uracil. Uracil is a pyrimidine, not a purine, and replaces thymine in RNA. Therefore, correct statements are A, B, D. Correct answer is option (1). Explanation length is exactly 50 words.
8) Which purine base pairs with cytosine in DNA?
(1) Adenine
(2) Guanine
(3) Uracil
(4) Thymine
Explanation: In DNA, guanine (a purine) pairs with cytosine (a pyrimidine) through three hydrogen bonds. Adenine pairs with thymine, and in RNA adenine pairs with uracil. Thus, the correct answer is guanine. Correct answer is option (2). Explanation length is exactly 50 words.
9) Which base pairing is correct in DNA?
(1) Adenine – Cytosine
(2) Adenine – Thymine
(3) Guanine – Adenine
(4) Cytosine – Thymine
Explanation: In DNA, complementary base pairing occurs such that adenine pairs with thymine by two hydrogen bonds, and guanine pairs with cytosine by three hydrogen bonds. Hence, the correct pairing among the options is adenine – thymine. Correct answer is option (2). Explanation length is exactly 50 words.
10) Which of the following are pyrimidines found in DNA?
(1) Adenine and Guanine
(2) Thymine and Cytosine
(3) Adenine and Cytosine
(4) Guanine and Thymine
Explanation: DNA contains two purines (adenine, guanine) and two pyrimidines (cytosine, thymine). Therefore, thymine and cytosine are the pyrimidines present in DNA. Correct answer is option (2). Explanation length is exactly 50 words.
Topic: Carbohydrates
Subtopic: Functional Groups of Sugars
Keyword Definitions:
• Sugars: Simple carbohydrates that serve as energy sources and structural molecules.
• Carbonyl group: Functional group with C=O bond, present in aldehydes and ketones.
• Hydroxyl group: Functional group with -OH, contributes to solubility and reactivity.
• Methyl group: Functional group with -CH3, not characteristic of sugars.
• Phosphate group: Functional group with PO4, present in nucleotides and some sugar derivatives.
• Functional group: Specific group of atoms responsible for chemical properties of a molecule.
Lead Question (2018):
The two functional groups characteristic of sugars are:
(A) Carbonyl and hydroxyl
(B) Hydroxyl and methyl
(C) Carbonyl and phosphate
(D) Carbonyl and methyl
Explanation:
The correct answer is (A) Carbonyl and hydroxyl. Sugars contain a carbonyl group (either aldehyde or ketone) and multiple hydroxyl groups (-OH). These functional groups define the chemical properties and reactivity of monosaccharides and disaccharides. Methyl and phosphate groups are not characteristic of normal sugars.
1. Monosaccharides with an aldehyde group are called:
(A) Ketoses
(B) Aldoses
(C) Polyols
(D) Disaccharides
Explanation:
The correct answer is (B) Aldoses. Monosaccharides containing an aldehyde carbonyl group are called aldoses. Ketoses contain a ketone group. These classifications are based on the position of the carbonyl functional group in the sugar molecule.
2. The general formula of monosaccharides is:
(A) CnH2nOn
(B) CnH2n-2On
(C) CnHnOn
(D) CnH2n-1On
Explanation:
The correct answer is (A) CnH2nOn. Monosaccharides follow the formula CnH2nOn, reflecting their composition of carbon, hydrogen, and oxygen. This general formula corresponds to sugars with both hydroxyl and carbonyl functional groups.
3. A ketose sugar has the carbonyl group at:
(A) Terminal carbon
(B) Second carbon
(C) Third carbon
(D) Any carbon
Explanation:
The correct answer is (B) Second carbon. Ketose sugars have a carbonyl (C=O) group at an internal carbon, usually C2. Aldoses have the carbonyl at the terminal carbon. Functional group position differentiates aldoses and ketoses.
4. Hydroxyl groups in sugars make them:
(A) Nonpolar
(B) Soluble in water
(C) Insoluble
(D) Reactive with lipids only
Explanation:
The correct answer is (B) Soluble in water. Hydroxyl (-OH) groups form hydrogen bonds with water, increasing solubility. They also participate in chemical reactions such as glycosidic bond formation. Carbonyl groups provide reactivity at the aldehyde or ketone site.
5. Reducing sugars are characterized by:
(A) Free carbonyl group
(B) Only hydroxyl groups
(C) Methyl groups
(D) Phosphate groups
Explanation:
The correct answer is (A) Free carbonyl group. Reducing sugars have a free aldehyde or ketone carbonyl group capable of reducing mild oxidizing agents like Benedict’s solution. Sugars without free carbonyl groups are non-reducing.
6. Glycosidic bonds are formed between:
(A) Carbonyl and hydroxyl
(B) Two carbonyl groups
(C) Two hydroxyl groups
(D) Hydroxyl and methyl
Explanation:
The correct answer is (C) Two hydroxyl groups. Glycosidic bonds link monosaccharides via condensation reactions between hydroxyl groups. Carbonyl groups remain available in reducing sugars for reactivity, but the bond itself forms between hydroxyl groups.
7. Assertion-Reason Question:
Assertion (A): Sugars are characterized by carbonyl and hydroxyl groups.
Reason (R): Carbonyl group reacts with hydroxyl groups to form cyclic structures.
(A) Both A and R true, R explains A
(B) Both A and R true, R does not explain A
(C) A true, R false
(D) A false, R true
Explanation:
Correct answer is (A). Sugars have both carbonyl and hydroxyl groups. The carbonyl group reacts intramolecularly with a hydroxyl to form cyclic hemiacetal or hemiketal structures. Both assertion and reason are true, and the reason explains the assertion.
8. Matching Type Question:
Match sugars with functional group:
(i) Glucose – (a) Aldose
(ii) Fructose – (b) Ketose
(iii) Ribose – (c) Aldose
(iv) Sucrose – (d) Disaccharide
(A) i-a, ii-b, iii-c, iv-d
(B) i-b, ii-a, iii-d, iv-c
(C) i-c, ii-d, iii-a, iv-b
(D) i-d, ii-c, iii-b, iv-a
Explanation:
Correct answer is (A). Glucose is an aldose, fructose a ketose, ribose an aldose monosaccharide, and sucrose is a disaccharide. Functional groups determine chemical reactivity and classification of sugars.
9. Fill in the Blanks:
The functional groups in sugars are ______ and ______.
(A) Hydroxyl, Methyl
(B) Carbonyl, Hydroxyl
(C) Carbonyl, Phosphate
(D) Carbonyl, Methyl
Explanation:
Correct answer is (B) Carbonyl, Hydroxyl. Sugars contain a carbonyl group (aldehyde or ketone) and multiple hydroxyl groups (-OH), which define their solubility and chemical reactivity. Methyl and phosphate groups are not typical of simple sugars.
10. Choose the correct statements:
(A) Sugars contain hydroxyl groups
(B) Sugars contain carbonyl groups
(C) Methyl groups are characteristic of sugars
(D) Phosphate is a typical sugar functional group
Options:
(1) A, B
(2) A, C
(3) B, D
(4) A, B, D
Explanation:
Correct answer is (1) A, B. Sugars are defined by hydroxyl and carbonyl groups. Methyl and phosphate groups are not characteristic of normal sugars. Understanding these functional groups is essential for NEET UG biomolecule questions.
Topic: DNA Analysis
Subtopic: Agarose Gel Electrophoresis
Keyword Definitions:
• DNA – Deoxyribonucleic acid, carrier of genetic information.
• Agarose gel – Porous matrix used for separation of nucleic acids.
• Ethidium bromide – Intercalating dye used to visualize DNA under UV light.
• Bromophenol blue – Tracking dye for monitoring electrophoresis progress.
• Acetocarmine – Stain primarily used for chromosomes in microscopy.
• Aniline blue – Stain for plant cell walls and callose.
• Gel electrophoresis – Technique to separate nucleic acids by size and charge.
• UV light – Ultraviolet light used to excite fluorescent dyes.
• DNA fragment – A portion of DNA separated by electrophoresis.
• Intercalating dye – Molecule that inserts between base pairs of DNA for visualization.
Lead Question – 2017:
The DNA fragments separated on an agarose gel can be visualised after staining with:
(A) Ethidium bromide
(B) Bromophenol blue
(C) Acetocarmine
(D) Aneline blue
Explanation:
DNA fragments on agarose gel are visualized using Ethidium bromide, which intercalates between base pairs and fluoresces under UV light. Bromophenol blue is a tracking dye, acetocarmine stains chromosomes, and aniline blue stains plant cell walls. (Answer: A)
1) Single Correct Answer MCQ:
Which dye is commonly used for visualizing RNA in gels?
(A) Ethidium bromide
(B) Coomassie blue
(C) Acetocarmine
(D) Silver nitrate
Explanation:
RNA in agarose gels can be visualized with Ethidium bromide, similar to DNA, due to its ability to intercalate between nucleic acid bases and fluoresce under UV light. (Answer: A)
2) Single Correct Answer MCQ:
Bromophenol blue in gel electrophoresis is used as:
(A) DNA stain
(B) Tracking dye
(C) Fixative
(D) Fluorescent marker
Explanation:
Bromophenol blue is used as a tracking dye to monitor the progress of DNA migration during electrophoresis. It does not stain DNA or fluoresce. (Answer: B)
3) Single Correct Answer MCQ:
Which light source is required to view DNA stained with Ethidium bromide?
(A) Visible light
(B) Infrared light
(C) UV light
(D) X-ray
Explanation:
UV light excites ethidium bromide intercalated in DNA, causing fluorescence and allowing visualization of DNA fragments on agarose gel. (Answer: C)
4) Single Correct Answer MCQ:
What is the role of agarose in electrophoresis?
(A) Staining
(B) Buffering
(C) Matrix for separation
(D) Tracking dye
Explanation:
Agarose forms a porous gel matrix through which DNA fragments migrate during electrophoresis, separating them based on size. It does not stain or act as a tracking dye. (Answer: C)
5) Single Correct Answer MCQ:
Which of the following is NOT used for nucleic acid visualization?
(A) Ethidium bromide
(B) Silver nitrate
(C) Acetocarmine
(D) SYBR Green
Explanation:
Acetocarmine is used to stain chromosomes, not nucleic acids in gels. Ethidium bromide, SYBR Green, and silver staining can be used to visualize DNA or RNA. (Answer: C)
6) Single Correct Answer MCQ:
Intercalating dyes like Ethidium bromide work by:
(A) Binding to sugar-phosphate backbone
(B) Inserting between base pairs
(C) Staining proteins
(D) Enhancing buffer conductivity
Explanation:
Ethidium bromide intercalates between DNA base pairs, allowing the DNA to fluoresce under UV light. This intercalation does not bind proteins or the sugar-phosphate backbone. (Answer: B)
7) Assertion-Reason MCQ:
Assertion (A): Ethidium bromide is mutagenic.
Reason (R): It intercalates between DNA bases and can cause frameshift mutations.
(A) Both A and R true, R explains A
(B) Both A and R true, R does not explain A
(C) A true, R false
(D) A false, R true
Explanation:
Both assertion and reason are correct. Ethidium bromide intercalates DNA and can cause frameshift mutations, making it mutagenic. (Answer: A)
8) Matching Type MCQ:
Match the dye with its use:
1. Ethidium bromide – (i) Protein staining
2. Coomassie blue – (ii) DNA visualization
3. Acetocarmine – (iii) Chromosome staining
4. Bromophenol blue – (iv) Tracking dye
Options:
(A) 1-ii, 2-i, 3-iii, 4-iv
(B) 1-i, 2-ii, 3-iii, 4-iv
(C) 1-iii, 2-i, 3-ii, 4-iv
(D) 1-ii, 2-iii, 3-i, 4-iv
Explanation:
Correct matching: Ethidium bromide for DNA (ii), Coomassie blue for protein (i), Acetocarmine for chromosomes (iii), Bromophenol blue as tracking dye (iv). (Answer: A)
9) Fill in the Blanks MCQ:
The fluorescent dye commonly used to visualize DNA under UV light is _______.
(A) Bromophenol blue
(B) Ethidium bromide
(C) Acetocarmine
(D) Aniline blue
Explanation:
Ethidium bromide is the standard fluorescent dye used in agarose gel electrophoresis to visualize DNA under UV illumination. (Answer: B)
10) Choose the correct statements MCQ:
1. Ethidium bromide intercalates DNA.
2. Bromophenol blue is used to stain DNA.
3. Acetocarmine stains chromosomes.
4. Aniline blue stains plant cell walls.
Options:
(A) 1, 3, 4
(B) 1, 2, 3
(C) 2, 3, 4
(D) 1, 2, 4
Explanation:
Statements 1, 3, 4 are correct. Bromophenol blue is a tracking dye, not a DNA stain. Ethidium bromide intercalates DNA, acetocarmine stains chromosomes, and aniline blue stains plant cell walls. (Answer: A)
Topic: Genetic Code and Protein Synthesis
Subtopic: Frameshift Mutation
Keyword Definitions:
• RNA – Ribonucleic acid, carries genetic information for protein synthesis.
• Codon – Sequence of three RNA bases coding for one amino acid.
• Amino acid – Building block of proteins.
• Protein – Macromolecule composed of amino acids, synthesized based on RNA code.
• Base deletion – Loss of a nucleotide in RNA or DNA sequence.
• Frameshift mutation – Mutation caused by insertion or deletion of nucleotides, altering reading frame.
• Position – Specific nucleotide location in the RNA sequence.
• Translation – Process of converting RNA codons into amino acids.
• Reading frame – Triplet grouping of nucleotides that determines codon translation.
• Clinical relevance – Frameshift mutations can lead to genetic diseases due to altered protein function.
Lead Question – 2017:
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
(A) 333
(B) 1
(C) 11
(D) 33
Explanation:
A deletion at position 901 causes a frameshift mutation starting at codon 301 (since 901 ÷ 3 = 300 codons complete, 301st codon affected). This frameshift alters all downstream codons until the stop codon, affecting 33 codons. Thus, 33 codons are altered due to the deletion. (Answer: D)
1) Single Correct Answer MCQ:
A mutation at the first base of an RNA coding sequence will affect:
(A) Only first amino acid
(B) Only last amino acid
(C) Entire protein
(D) No amino acids
Explanation:
A deletion at the first base causes a frameshift mutation, altering the entire reading frame. Consequently, all downstream codons are changed, affecting the entire protein sequence. Only first amino acid alone is not affected; the shift continues through the RNA. (Answer: C)
2) Single Correct Answer MCQ:
A codon is made up of:
(A) 2 nucleotides
(B) 3 nucleotides
(C) 4 nucleotides
(D) 5 nucleotides
Explanation:
A codon consists of three consecutive nucleotides in RNA. Each codon codes for a specific amino acid, ensuring correct translation. Two or more nucleotides would not suffice to code for all 20 amino acids. (Answer: B)
3) Single Correct Answer MCQ:
Frameshift mutations are typically caused by:
(A) Substitution
(B) Deletion or insertion
(C) Methylation
(D) Splicing
Explanation:
Frameshift mutations occur due to deletion or insertion of nucleotides not in multiples of three, which shifts the reading frame and changes all downstream codons. Substitutions affect a single codon, while methylation or splicing do not shift the reading frame. (Answer: B)
4) Single Correct Answer MCQ:
Which amino acid is coded by AUG?
(A) Methionine
(B) Serine
(C) Valine
(D) Leucine
Explanation:
The AUG codon is the start codon and codes for methionine. It signals the beginning of translation and ensures correct initiation of protein synthesis. Other codons like serine or valine are coded by different sequences. (Answer: A)
5) Single Correct Answer MCQ:
A deletion at the last nucleotide of RNA affects:
(A) Only last codon
(B) All codons
(C) First codon
(D) No codons
Explanation:
A deletion at the last nucleotide alters only the last codon because upstream codons remain unaffected. Frameshift only affects downstream codons, so previous codons translate normally. (Answer: A)
6) Single Correct Answer MCQ:
Which type of mutation is least likely to affect protein sequence?
(A) Silent mutation
(B) Frameshift
(C) Nonsense
(D) Insertion
Explanation:
A silent mutation changes a nucleotide without altering the amino acid sequence due to codon degeneracy. Frameshifts, nonsense, and insertions generally alter the protein sequence, potentially causing nonfunctional proteins. (Answer: A)
7) Assertion-Reason MCQ:
Assertion (A): Deleting one nucleotide can alter many codons.
Reason (R): RNA is read in triplets called codons.
(A) Both A and R true, R is correct explanation
(B) Both A and R true, R not correct explanation
(C) A true, R false
(D) A false, R true
Explanation:
Both A and R are true. The RNA is read in triplets; a deletion shifts the reading frame, changing all downstream codons. The reading frame principle explains why a single deletion can affect multiple codons. (Answer: A)
8) Matching Type MCQ:
Match mutation type with effect:
(A) Silent – (i) No amino acid change
(B) Nonsense – (ii) Premature stop codon
(C) Frameshift – (iii) Alters downstream codons
(D) Missense – (iv) Changes single amino acid
Options:
(A) A-i, B-ii, C-iii, D-iv
(B) A-ii, B-iii, C-i, D-iv
(C) A-iii, B-i, C-iv, D-ii
(D) A-iv, B-i, C-ii, D-iii
Explanation:
Correct matching: Silent mutations do not change amino acids, nonsense introduces premature stop codons, frameshift alters all downstream codons, and missense changes one amino acid. (Answer: A)
9) Fill in the Blanks MCQ:
A deletion or insertion of nucleotides not in multiples of three results in ________ mutation.
(A) Frameshift
(B) Silent
(C) Missense
(D) Nonsense
Explanation:
Deletion or insertion of nucleotides not divisible by three causes a frameshift mutation, shifting the reading frame and altering all downstream codons. Silent mutations do not affect codons, and missense/nonsense affect individual codons. (Answer: A)
10) Choose the correct statements MCQ:
1. Each codon codes for one amino acid.
2. A frameshift mutation can alter multiple amino acids.
3. Deleting three nucleotides may not cause frameshift.
4. AUG is a start codon.
Options:
(A) 1, 2, 3, 4
(B) 1, 2 only
(C) 2, 3, 4 only
(D) 1, 4 only
Explanation:
All statements 1, 2, 3, 4 are correct. Codons are triplets; frameshifts alter downstream amino acids; deletion of three nucleotides preserves reading frame; AUG is start codon. This knowledge is crucial for understanding mutation effects on proteins. (Answer: A)
Topic: DNA Replication
Subtopic: Okazaki Fragments and Strand Elongation
Keyword Definitions:
• DNA replication – Process of copying DNA prior to cell division.
• Leading strand – DNA strand synthesized continuously toward replication fork.
• Lagging strand – DNA strand synthesized discontinuously away from replication fork.
• Okazaki fragments – Short DNA segments synthesized on lagging strand.
• DNA polymerase – Enzyme that synthesizes DNA by adding nucleotides.
• Replication fork – Y-shaped structure where DNA is unwound for replication.
• RNA primer – Short RNA sequence required to initiate DNA synthesis.
• Clinical relevance – Errors in Okazaki fragment processing can lead to mutations or genomic instability.
Lead Question – 2017:
During DNA replication, Okazaki fragments are used to elongate :
(A) The lagging strand away from the replication fork
(B) The leading strand towards replication fork
(C) The lagging strand towards replication fork
(D) The leading strand away from replication fork
Explanation:
Okazaki fragments are short DNA sequences synthesized discontinuously on the lagging strand away from the replication fork. DNA polymerase extends these fragments in the 5'→3' direction, while ligase joins them. This allows complete replication of the lagging strand. (Answer: A)
1) Which enzyme joins Okazaki fragments?
(A) DNA ligase
(B) DNA polymerase
(C) Helicase
(D) Primase
Explanation:
DNA ligase seals gaps between Okazaki fragments on the lagging strand, creating a continuous DNA strand. Polymerase synthesizes fragments, primase lays RNA primers, and helicase unwinds DNA. Proper ligation is essential for genomic stability. (Answer: A)
2) The leading strand is synthesized:
(A) Continuously towards replication fork
(B) Discontinuously away from replication fork
(C) Continuously away from replication fork
(D) Discontinuously towards replication fork
Explanation:
The leading strand is synthesized continuously in the 5'→3' direction toward the replication fork, unlike the lagging strand, which is discontinuous. This ensures efficient and uninterrupted DNA synthesis on one strand. (Answer: A)
3) RNA primers are necessary for:
(A) Initiation of Okazaki fragment synthesis
(B) Sealing DNA gaps
(C) DNA unwinding
(D) Proofreading
Explanation:
RNA primers provide a free 3'-OH end for DNA polymerase to start synthesis of Okazaki fragments. They are removed later and replaced with DNA. Primers are crucial for initiating lagging strand replication. (Answer: A)
4) Helicase function in replication:
(A) Unwinds DNA double helix
(B) Synthesizes RNA primers
(C) Joins Okazaki fragments
(D) Proofreads DNA
Explanation:
Helicase unwinds the DNA double helix at the replication fork, allowing polymerase to access single strands. It does not join fragments or proofread. This is essential for replication fork progression. (Answer: A)
5) Which strand requires multiple primers?
(A) Lagging strand
(B) Leading strand
(C) Both strands
(D) Neither strand
Explanation:
The lagging strand is synthesized in short Okazaki fragments, each requiring a new RNA primer. The leading strand uses a single primer for continuous synthesis. Multiple primers enable discontinuous replication. (Answer: A)
6) Clinical relevance of Okazaki fragment errors:
(A) Mutations
(B) Genomic instability
(C) DNA repair defects
(D) All of the above
Explanation:
Improper synthesis or ligation of Okazaki fragments can cause mutations, genomic instability, and DNA repair defects, contributing to cancer or genetic disorders. Accurate replication is critical for cell viability. (Answer: D)
7) Assertion-Reason type:
Assertion (A): Okazaki fragments elongate the lagging strand.
Reason (R): DNA polymerase can synthesize DNA only in 5'→3' direction.
(A) Both A and R true, R is correct explanation
(B) Both A and R true, R not correct explanation
(C) A true, R false
(D) A false, R true
Explanation:
Both A and R are true. DNA polymerase synthesizes DNA only in the 5'→3' direction, requiring the lagging strand to be made discontinuously as Okazaki fragments away from the replication fork. (Answer: A)
8) Matching type:
Match:
(A) DNA polymerase – (i) Synthesizes DNA
(B) Primase – (ii) Lays RNA primer
(C) Ligase – (iii) Joins DNA fragments
(D) Helicase – (iv) Unwinds DNA
Options:
(A) A-i, B-ii, C-iii, D-iv
(B) A-ii, B-i, C-iv, D-iii
(C) A-iii, B-iv, C-ii, D-i
(D) A-iv, B-iii, C-i, D-ii
Explanation:
Correct match: DNA polymerase synthesizes DNA, primase lays RNA primers, ligase joins Okazaki fragments, helicase unwinds DNA. This coordination ensures efficient replication of both strands. (Answer: A)
9) Fill in the blanks:
Short DNA sequences on the lagging strand are called ________.
(A) Okazaki fragments
(B) Primers
(C) Telomeres
(D) Exons
Explanation:
Discontinuous DNA segments synthesized on the lagging strand are Okazaki fragments. They are later joined by ligase to form a continuous strand. This is a key feature of DNA replication. (Answer: A)
10) Choose the correct statements:
1. Lagging strand is synthesized discontinuously.
2. Leading strand requires multiple primers.
3. Ligase joins Okazaki fragments.
4. Helicase unwinds DNA.
Options:
(A) 1, 3, 4 only
(B) 1, 2, 3 only
(C) 2, 3, 4 only
(D) All correct
Explanation:
Statements 1, 3, and 4 are correct. The lagging strand is discontinuous, ligase joins fragments, and helicase unwinds DNA. The leading strand requires only a single primer for continuous synthesis. (Answer: A)
Topic: RNA and Protein Synthesis
Subtopic: Types of RNA in Animal Cells
Keyword Definitions:
• RNA – Ribonucleic acid, molecule involved in protein synthesis and regulation.
• rRNA – Ribosomal RNA, structural and functional component of ribosomes.
• tRNA – Transfer RNA, carries amino acids to ribosomes for translation.
• mRNA – Messenger RNA, carries coding information from DNA to ribosomes.
• miRNA – Micro RNA, regulates gene expression post-transcriptionally.
• Ribosome – Cellular machinery for protein synthesis, composed of rRNA and proteins.
• Abundance – Relative quantity of each RNA type in the cell.
• Clinical relevance – rRNA and tRNA levels affect protein synthesis efficiency; miRNA dysregulation may cause disease.
Lead Question – 2017:
Which of the following RNAs should be most abundant in animal cell?
(A) mi-RNA
(B) r-RNA
(C) t-RNA
(D) m-RNA
Explanation:
Ribosomal RNA (rRNA) is most abundant in animal cells because it forms the core of ribosomes, which are required in large numbers for protein synthesis. tRNA and mRNA are less abundant, and miRNA exists in small regulatory amounts. High rRNA abundance ensures efficient translation. (Answer: B)
1) Which RNA carries amino acids to ribosomes?
(A) tRNA
(B) rRNA
(C) mRNA
(D) miRNA
Explanation:
Transfer RNA (tRNA) delivers specific amino acids to ribosomes during translation. Each tRNA recognizes codons on mRNA, ensuring proper protein assembly. This functional specificity distinguishes tRNA from rRNA or miRNA. (Answer: A)
2) The RNA type that encodes protein sequences is:
(A) mRNA
(B) rRNA
(C) tRNA
(D) miRNA
Explanation:
Messenger RNA (mRNA) carries genetic information transcribed from DNA to ribosomes for protein synthesis. It determines amino acid sequence, unlike rRNA or tRNA, which are structural or transport RNAs, and miRNA, which regulates expression. (Answer: A)
3) MicroRNAs function to:
(A) Regulate gene expression
(B) Form ribosomes
(C) Carry amino acids
(D) Encode proteins
Explanation:
miRNAs are small RNAs that bind mRNA transcripts and suppress translation or promote degradation. They do not encode proteins, form ribosomes, or transport amino acids, but are crucial for post-transcriptional regulation and cellular homeostasis. (Answer: A)
4) Ribosomes consist primarily of:
(A) rRNA and proteins
(B) mRNA and proteins
(C) tRNA and lipids
(D) miRNA and DNA
Explanation:
Ribosomes are made of rRNA and proteins, forming two subunits. rRNA catalyzes peptide bond formation, while proteins stabilize structure. mRNA and tRNA interact with ribosomes but are not structural components. (Answer: A)
5) Which RNA is least abundant in a typical animal cell?
(A) miRNA
(B) rRNA
(C) tRNA
(D) mRNA
Explanation:
MicroRNA (miRNA) is least abundant, present in nanomolar amounts, as it regulates specific genes post-transcriptionally. In contrast, rRNA is most abundant for ribosome structure, and tRNA and mRNA exist in intermediate amounts. (Answer: A)
6) Clinical relevance of RNA abundance:
(A) High rRNA ensures protein synthesis
(B) Low miRNA may deregulate genes
(C) tRNA levels affect translation efficiency
(D) All of the above
Explanation:
All statements are correct. High rRNA supports ribosome function, tRNA abundance affects translation, and miRNA dysregulation can lead to diseases like cancer. RNA balance is crucial for cellular health and protein production. (Answer: D)
7) Assertion-Reason type:
Assertion (A): rRNA is most abundant in cells.
Reason (R): Ribosomes are needed in large numbers for protein synthesis.
(A) Both A and R true, R is correct explanation
(B) Both A and R true, R not correct explanation
(C) A true, R false
(D) A false, R true
Explanation:
Both assertion and reason are true; rRNA is abundant because cells require numerous ribosomes for protein synthesis. High rRNA ensures efficient translation in metabolically active cells. (Answer: A)
8) Matching type:
Match RNA with function:
(A) rRNA – (i) Structural and catalytic component of ribosomes
(B) tRNA – (ii) Delivers amino acids
(C) mRNA – (iii) Encodes protein
(D) miRNA – (iv) Gene expression regulation
Options:
(A) A-i, B-ii, C-iii, D-iv
(B) A-ii, B-i, C-iv, D-iii
(C) A-iii, B-iv, C-ii, D-i
(D) A-iv, B-iii, C-i, D-ii
Explanation:
Correct match: rRNA – structural/catalytic, tRNA – amino acid transport, mRNA – encodes protein, miRNA – regulates gene expression. This classification clarifies RNA roles in animal cells. (Answer: A)
9) Fill in the blanks:
The RNA type forming the catalytic core of ribosomes is ________.
(A) rRNA
(B) tRNA
(C) mRNA
(D) miRNA
Explanation:
Ribosomal RNA (rRNA) forms the catalytic and structural core of ribosomes. It ensures peptide bond formation during translation. tRNA, mRNA, and miRNA serve transport, coding, or regulatory functions. (Answer: A)
10) Choose the correct statements:
1. rRNA is most abundant in cells.
2. mRNA is less abundant than rRNA.
3. tRNA is required for translation.
4. miRNA is highly abundant.
Options:
(A) 1, 2, 3 only
(B) 1, 3, 4 only
(C) 2, 3, 4 only
(D) All correct
Explanation:
Statements 1, 2, and 3 are correct; rRNA is most abundant, mRNA is intermediate, tRNA participates in translation. miRNA is not highly abundant but acts as regulatory RNA in small amounts. (Answer: A)
Topic: Biological Macromolecules
Subtopic: Polymers and Non-Polymers
Keyword Definitions:
• Polymer – Large molecule formed by repeating structural units (monomers).
• Monomer – Small unit that joins to form a polymer.
• Lipids – Hydrophobic molecules, non-polymeric, include fats, oils, steroids.
• Proteins – Polymers of amino acids, perform structural and catalytic roles.
• Nucleic acids – Polymers of nucleotides, carry genetic information.
• Polysaccharides – Polymers of monosaccharides, energy storage and structural molecules.
• Clinical relevance – Storage disorders (glycogen storage disease) or lipid metabolism disorders affect health.
Lead Question – 2017:
Which of the following are not polymeric ?
(A) Lipids
(B) Nucleic acids
(C) Proteins
(D) Polysaccharides
Explanation:
Among the given biomolecules, lipids are not polymeric because they are not formed by repeating monomeric units. Instead, lipids are heterogeneous molecules such as triglycerides and steroids. Nucleic acids, proteins, and polysaccharides are true polymers formed by nucleotide, amino acid, and monosaccharide monomers respectively. (Answer: A)
1) Which of the following is a storage polysaccharide in animals?
(A) Cellulose
(B) Glycogen
(C) Starch
(D) Chitin
Explanation:
Glycogen is the primary storage polysaccharide in animals, mainly stored in liver and muscle. It is a branched polymer of glucose. Glycogen metabolism is tightly regulated by insulin and glucagon. Disorders like glycogen storage disease occur due to enzyme deficiencies. (Answer: B)
2) Which monomer forms proteins?
(A) Amino acids
(B) Nucleotides
(C) Fatty acids
(D) Monosaccharides
Explanation:
Proteins are polymers of amino acids, linked through peptide bonds. The sequence of amino acids determines protein structure and function. Mutations in amino acid sequence can cause structural abnormalities leading to clinical disorders such as sickle cell anemia. (Answer: A)
3) Clinical case: A child presents with hepatomegaly and hypoglycemia due to deficiency of glucose-6-phosphatase. This condition is:
(A) Tay-Sachs disease
(B) Gaucher’s disease
(C) Von Gierke’s disease
(D) Pompe’s disease
Explanation:
Von Gierke’s disease is a glycogen storage disorder due to glucose-6-phosphatase deficiency. It leads to accumulation of glycogen in the liver, causing hepatomegaly and hypoglycemia. Proper dietary management reduces complications. It highlights clinical significance of glycogen metabolism. (Answer: C)
4) Which bond joins monosaccharides in polysaccharides?
(A) Peptide bond
(B) Glycosidic bond
(C) Ester bond
(D) Phosphodiester bond
Explanation:
Monosaccharides are joined together in polysaccharides by glycosidic bonds. This covalent bond forms between the hydroxyl groups of monosaccharides, releasing water. Glycosidic linkages determine polysaccharide structure and digestibility, such as α-linkage in starch versus β-linkage in cellulose. (Answer: B)
5) Which of the following is not a carbohydrate?
(A) Glucose
(B) Fructose
(C) Glycerol
(D) Galactose
Explanation:
Glycerol is not a carbohydrate but a component of lipids. It combines with fatty acids to form triglycerides. In contrast, glucose, fructose, and galactose are monosaccharides that function as primary energy sources. (Answer: C)
6) The genetic material in most organisms is:
(A) Protein
(B) DNA
(C) RNA
(D) Lipid
Explanation:
In most organisms, DNA serves as the genetic material, encoding instructions for protein synthesis and inheritance. RNA acts as the genetic material in some viruses. DNA’s polymeric nucleotide structure ensures stable storage of genetic information. (Answer: B)
7) Assertion-Reason type:
Assertion (A): Cellulose cannot be digested by humans.
Reason (R): Humans lack the enzyme cellulase.
(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is not the correct explanation
(C) A is true, R is false
(D) A is false, R is true
Explanation:
Both assertion and reason are true, and R explains A. Humans lack cellulase, the enzyme required to hydrolyze β-1,4-glycosidic bonds in cellulose. Thus, cellulose passes undigested, acting as dietary fiber, important for bowel health. (Answer: A)
8) Matching type:
Match biomolecules with their monomers:
(A) Proteins – (i) Nucleotides
(B) Nucleic acids – (ii) Amino acids
(C) Polysaccharides – (iii) Monosaccharides
(D) Lipids – (iv) Glycerol + Fatty acids
Options:
(A) A-ii, B-i, C-iii, D-iv
(B) A-i, B-ii, C-iv, D-iii
(C) A-iii, B-iv, C-i, D-ii
(D) A-iv, B-iii, C-ii, D-i
Explanation:
Proteins are formed from amino acids, nucleic acids from nucleotides, polysaccharides from monosaccharides, and lipids from glycerol plus fatty acids. These basic units determine the structure and function of biomolecules. (Answer: A)
9) Fill in the blanks:
The polysaccharide present in the exoskeleton of insects is ______.
(A) Chitin
(B) Cellulose
(C) Starch
(D) Glycogen
Explanation:
Chitin is the structural polysaccharide in arthropod exoskeletons. It consists of N-acetylglucosamine units with β-1,4 linkages. It provides rigidity and protection. Chitin is also found in fungal cell walls, serving structural roles. (Answer: A)
10) Choose the correct statements:
1. Lipids are non-polymeric molecules.
2. Proteins are made up of amino acids.
3. DNA is a polymer of nucleotides.
4. Polysaccharides are polymers of fatty acids.
(A) 1, 2, 3 only
(B) 2 and 4 only
(C) 1 and 3 only
(D) All are correct
Explanation:
Statements 1, 2, and 3 are correct. Statement 4 is incorrect because polysaccharides are polymers of monosaccharides, not fatty acids. This classification is vital in understanding biomolecular structure and function. Correct answer: 1, 2, 3 only. (Answer: A)
Topic: Enzymes
Subtopic: Structure and Components of Enzymes
Keyword Definitions:
• Enzyme – Biological catalyst that speeds up biochemical reactions.
• Apoenzyme – Protein portion of an enzyme, inactive without non-protein part.
• Coenzyme – Organic, non-protein cofactor often derived from vitamins.
• Cofactor – Non-protein component essential for enzyme activity (can be metal ions or coenzymes).
• Holoenzyme – Complete, active enzyme composed of apoenzyme and cofactor/coenzyme.
• Clinical relevance – Enzyme deficiencies cause metabolic disorders like phenylketonuria, albinism, or SCID.
Lead Question – 2017:
Which one of the following statements is correct, with reference to enzymes ?
(A) Holoenzyme = Coenzyme + Co-factor
(B) Apoenzyme = Holoenzyme + Coenzyme
(C) Holoenzyme = Apoenzyme + Coenzyme
(D) Coenzyme = Apoenzyme + Holoenzyme
Explanation:
The correct statement is Holoenzyme = Apoenzyme + Coenzyme. Apoenzyme alone is inactive, and activity is achieved only after binding with coenzyme or other cofactors, forming the holoenzyme. This complete enzyme participates effectively in biochemical reactions, ensuring proper metabolism. This principle is clinically important in understanding vitamin deficiencies. (Answer: C)
1) The enzyme that catalyzes conversion of glucose to glucose-6-phosphate is:
(A) Hexokinase
(B) Glucokinase
(C) Phosphatase
(D) Aldolase
Explanation:
Hexokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate, the first step of glycolysis. This reaction requires ATP and magnesium as cofactors. The process traps glucose within the cell and initiates energy production. Defects in hexokinase can impair glycolysis and glucose metabolism. (Answer: A)
2) Which vitamin-derived coenzyme is essential for transamination reactions?
(A) Thiamine pyrophosphate
(B) Pyridoxal phosphate
(C) FAD
(D) NAD+
Explanation:
Pyridoxal phosphate (vitamin B6 derivative) acts as a coenzyme in transamination reactions, transferring amino groups between amino acids and keto acids. It is crucial in amino acid metabolism. Clinical deficiency leads to seizures, neuropathy, and impaired neurotransmitter synthesis. (Answer: B)
3) Clinical case: A child presents with developmental delay and mousy odor urine. Deficiency of which enzyme is implicated?
(A) Hexokinase
(B) Phenylalanine hydroxylase
(C) Lactase
(D) Catalase
Explanation:
Phenylketonuria (PKU) is caused by deficiency of phenylalanine hydroxylase. This leads to accumulation of phenylalanine and its metabolites, causing neurotoxicity, seizures, and intellectual disability. Early detection and dietary restriction of phenylalanine prevent irreversible complications. This clinical example shows enzyme deficiency impact. (Answer: B)
4) Enzyme activity is measured in:
(A) Grams
(B) Katal
(C) Meters
(D) Joules
Explanation:
Enzyme activity is expressed in Katal, defined as the amount of enzyme converting one mole of substrate per second. Clinically, enzyme activity assays help diagnose myocardial infarction (CK-MB, LDH) and liver disorders (AST, ALT, ALP). This measurement is crucial in biochemical and clinical laboratories. (Answer: B)
5) Enzymes are mainly composed of:
(A) Lipids
(B) Carbohydrates
(C) Proteins
(D) Nucleic acids
Explanation:
Most enzymes are proteins in nature, composed of amino acids folded into specific structures that determine their catalytic activity. Ribozymes are exceptions, being RNA enzymes. Enzyme dysfunction due to misfolding or genetic mutations leads to various clinical diseases. (Answer: C)
6) Which factor does not affect enzyme activity?
(A) Temperature
(B) pH
(C) Substrate concentration
(D) Magnetic field
Explanation:
Enzyme activity is influenced by temperature, pH, and substrate concentration, but not directly by magnetic fields. Deviations from optimal temperature or pH denature enzymes, reducing their activity. This knowledge is used in therapeutic enzyme storage and industrial enzyme applications. (Answer: D)
7) Assertion-Reason type:
Assertion (A): Enzyme catalysis lowers activation energy.
Reason (R): Enzymes provide an alternate reaction pathway.
(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is not the correct explanation
(C) A is true, R is false
(D) A is false, R is true
Explanation:
Both assertion and reason are true, and R explains A. Enzymes reduce activation energy by stabilizing transition states, thereby accelerating reactions. They provide an alternate pathway without being consumed. This principle explains how enzymes achieve remarkable reaction rate enhancements in biology. (Answer: A)
8) Matching type:
Match the enzyme with its coenzyme:
(A) Lactate dehydrogenase – (i) NAD+
(B) Pyruvate dehydrogenase – (ii) Thiamine pyrophosphate
(C) Transaminase – (iii) Pyridoxal phosphate
(D) Succinate dehydrogenase – (iv) FAD
Options:
(A) A-i, B-ii, C-iii, D-iv
(B) A-ii, B-i, C-iv, D-iii
(C) A-iii, B-iv, C-ii, D-i
(D) A-i, B-iii, C-ii, D-iv
Explanation:
Correct matching: Lactate dehydrogenase – NAD+, Pyruvate dehydrogenase – Thiamine pyrophosphate, Transaminase – Pyridoxal phosphate, Succinate dehydrogenase – FAD. These enzyme-coenzyme relationships are critical in metabolic pathways. Deficiencies of respective vitamins impair enzymatic function, causing metabolic diseases. (Answer: A)
9) Fill in the blanks:
The enzyme that breaks down starch into maltose is ______.
(A) Amylase
(B) Lipase
(C) Lactase
(D) Sucrase
Explanation:
Amylase, secreted by salivary glands and pancreas, hydrolyzes starch into maltose. It initiates carbohydrate digestion in the mouth and continues in the small intestine. Amylase measurement is clinically important in diagnosing acute pancreatitis. This enzyme deficiency can impair carbohydrate metabolism. (Answer: A)
10) Choose the correct statements:
1. Enzymes are protein catalysts.
2. Holoenzyme = Apoenzyme + Cofactor.
3. Vitamins act as precursors for many coenzymes.
4. Enzyme catalysis consumes the enzyme.
(A) 1, 2, 3 only
(B) 1 and 4 only
(C) 2 and 3 only
(D) All are correct
Explanation:
Statements 1, 2, and 3 are correct. Statement 4 is incorrect because enzymes are not consumed during catalysis and remain available for reuse. This feature ensures high efficiency of enzymatic reactions in biological systems. Correct answer: 1, 2, 3 only. (Answer: A)
Topic: DNA Structure and Properties
Subtopic: Charge of DNA Fragments
Keyword Definitions:
• DNA fragments – Short pieces of DNA generated during restriction digestion or PCR.
• Negatively charged – Property due to phosphate groups in the DNA backbone.
• Electrophoresis – Laboratory method to separate charged molecules in a gel under electric field.
• Clinical relevance – Used in genetic diagnosis, forensic science, and molecular medicine.
Lead Question – 2017:
DNA fragments are :
(A) Either positively or negatively charged depending on their size
(B) Positively charged
(C) Negatively charged
(D) Neutral
Explanation:
DNA fragments are negatively charged due to phosphate groups in their sugar-phosphate backbone. This negative charge is independent of fragment size and allows separation through agarose gel electrophoresis toward the positive electrode. This principle is widely used in clinical molecular diagnostics, DNA fingerprinting, and genetic research applications. (Answer: C)
1) DNA fingerprinting uses:
(A) DNA ligase
(B) DNA polymerase
(C) Restriction endonuclease
(D) RNA polymerase
Explanation:
DNA fingerprinting uses restriction endonucleases to cut DNA into fragments at specific sequences. These fragments are then separated by electrophoresis and hybridized with probes. This method helps in criminal investigations, paternity testing, and genetic diagnosis. Its accuracy depends on polymorphisms in noncoding regions. (Answer: C)
2) The enzyme used in PCR is:
(A) DNA polymerase I
(B) DNA polymerase III
(C) Taq polymerase
(D) RNA polymerase
Explanation:
PCR uses Taq polymerase, a heat-stable enzyme isolated from Thermus aquaticus. It withstands high denaturation temperatures during PCR cycles. This allows amplification of DNA sequences for medical diagnostics, forensic studies, and genetic analysis. The enzyme synthesizes new DNA strands using primers. (Answer: C)
3) In gel electrophoresis, DNA fragments move toward:
(A) Cathode
(B) Anode
(C) Remain stationary
(D) Random directions
Explanation:
In gel electrophoresis, DNA fragments migrate toward the anode because DNA is negatively charged. The separation depends on fragment size, with smaller fragments moving faster through the gel matrix. This technique is fundamental in molecular biology, clinical diagnostics, and DNA profiling. (Answer: B)
4) Clinical case: A newborn is suspected of cystic fibrosis. Which molecular test confirms diagnosis?
(A) Southern blotting
(B) ELISA
(C) PCR
(D) Western blotting
Explanation:
For cystic fibrosis diagnosis, PCR is used to amplify and detect mutations in the CFTR gene. PCR provides rapid and accurate results from small DNA samples. It is routinely used in clinical genetics for early detection of inherited disorders and carrier screening. (Answer: C)
5) Okazaki fragments are synthesized during:
(A) Leading strand synthesis
(B) Lagging strand synthesis
(C) RNA transcription
(D) Protein translation
Explanation:
Okazaki fragments are short DNA pieces synthesized on the lagging strand during DNA replication. DNA ligase later joins them to form a continuous strand. This process ensures accurate DNA duplication and prevents genetic instability. They are essential for semi-discontinuous replication. (Answer: B)
6) Which bond stabilizes DNA double helix?
(A) Ionic bonds
(B) Hydrogen bonds
(C) Peptide bonds
(D) Disulfide bonds
Explanation:
DNA double helix stability is provided by hydrogen bonds between complementary nitrogenous bases (A–T and G–C). These noncovalent interactions, along with base stacking, ensure structural integrity and accurate genetic information transfer during replication and transcription. (Answer: B)
7) Assertion-Reason type:
Assertion (A): DNA is negatively charged.
Reason (R): DNA contains phosphate groups in its backbone.
(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is not the correct explanation
(C) A is true, R is false
(D) A is false, R is true
Explanation:
Both assertion and reason are correct. DNA is negatively charged because phosphate groups in its sugar-phosphate backbone carry negative charges. This property is fundamental for electrophoresis and molecular diagnostics. Hence, R correctly explains A. (Answer: A)
8) Matching type:
Match the following enzymes with their functions:
(A) DNA polymerase – (i) Joins Okazaki fragments
(B) Ligase – (ii) DNA synthesis
(C) Helicase – (iii) Unwinds DNA
(D) Topoisomerase – (iv) Relieves supercoiling
Options:
(A) A-ii, B-i, C-iii, D-iv
(B) A-i, B-ii, C-iii, D-iv
(C) A-iv, B-i, C-ii, D-iii
(D) A-ii, B-iii, C-i, D-iv
Explanation:
DNA polymerase synthesizes new DNA (ii), ligase joins Okazaki fragments (i), helicase unwinds DNA (iii), and topoisomerase relieves supercoiling (iv). Correct matching: A-ii, B-i, C-iii, D-iv. These enzymes act together during DNA replication, ensuring accuracy and continuity. (Answer: A)
9) Fill in the blanks:
During DNA replication, primers are made of ______.
(A) DNA
(B) RNA
(C) Protein
(D) Carbohydrates
Explanation:
During DNA replication, primers are made of RNA, synthesized by primase enzyme. These short RNA sequences provide a free 3′-OH group for DNA polymerase to initiate DNA synthesis. Later, RNA primers are replaced with DNA for continuity. (Answer: B)
10) Choose the correct statements:
1. DNA is double-stranded and negatively charged.
2. RNA is usually single-stranded.
3. DNA replication is semi-conservative.
4. Okazaki fragments are on the leading strand.
(A) 1, 2, 3 only
(B) 2 and 4 only
(C) 1 and 3 only
(D) All are correct
Explanation:
Statements 1, 2, and 3 are correct. Statement 4 is incorrect because Okazaki fragments occur on the lagging strand, not leading. Thus, the correct answer is 1, 2, and 3 only. These features are central to molecular biology and genetics. (Answer: A)
Chapter: Molecular Biology
Topic: RNA Structure and Function
Subtopic: Ribosomal RNA and Ribozymes
Keyword Definitions:
• rRNA: Ribosomal RNA, a structural and functional component of ribosomes.
• Ribozyme: RNA molecule with catalytic activity.
• Structural RNA: RNA that forms the framework of ribosomes and other complexes.
• 5.8 S rRNA: Component of eukaryotic large ribosomal subunit.
• 5 S rRNA: Small rRNA of the large subunit, structural role.
• 18 S rRNA: Component of the small subunit in eukaryotic ribosomes.
• 23 S rRNA: Component of the large subunit in bacterial ribosomes; acts as structural RNA and ribozyme.
Lead Question - 2016 (Phase 2):
Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?
(1) 5.8 S rRNA
(2) 5 S rRNA
(3) 18 S rRNA
(4) 23 S rRNA
Explanation: In bacteria, 23 S rRNA is part of the large ribosomal subunit, providing structural support and catalyzing peptide bond formation during translation, functioning as a ribozyme. Correct answer: (4) 23 S rRNA.
1. Single Correct Answer MCQ:
Which bacterial rRNA catalyzes peptide bond formation?
(1) 16 S rRNA
(2) 23 S rRNA
(3) 5 S rRNA
(4) tRNA
Explanation: 23 S rRNA of the bacterial large ribosomal subunit catalyzes peptide bond formation, demonstrating ribozyme activity and ensuring protein synthesis. Correct answer: (2) 23 S rRNA.
2. Single Correct Answer MCQ:
Which rRNA is part of the bacterial small ribosomal subunit?
(1) 16 S rRNA
(2) 23 S rRNA
(3) 5.8 S rRNA
(4) 18 S rRNA
Explanation: 16 S rRNA is part of the 30S small subunit in bacteria and is essential for mRNA decoding and ribosome assembly. Correct answer: (1) 16 S rRNA.
3. Single Correct Answer MCQ (Clinical-type):
Mutation in 23 S rRNA in bacteria affects:
(1) DNA replication
(2) Transcription
(3) Protein synthesis
(4) Lipid synthesis
Explanation: Mutations in 23 S rRNA disrupt ribosomal structure and catalytic activity, impairing peptide bond formation and protein synthesis, which can lead to antibiotic resistance. Correct answer: (3) Protein synthesis.
4. Single Correct Answer MCQ:
Which rRNA is part of eukaryotic 60S subunit equivalent to bacterial 23 S?
(1) 5.8 S rRNA
(2) 18 S rRNA
(3) 5 S rRNA
(4) 28 S rRNA
Explanation: Eukaryotic 28 S rRNA is functionally similar to bacterial 23 S rRNA, providing structural support and catalyzing peptide bond formation in the large ribosomal subunit. Correct answer: (4) 28 S rRNA.
5. Single Correct Answer MCQ:
Which rRNA is targeted by macrolide antibiotics in bacteria?
(1) 16 S rRNA
(2) 23 S rRNA
(3) tRNA
(4) 5 S rRNA
Explanation: Macrolides bind to 23 S rRNA in the 50S bacterial ribosomal subunit, blocking the exit tunnel and inhibiting protein synthesis. Correct answer: (2) 23 S rRNA.
6. Single Correct Answer MCQ:
Which rRNA has both structural and catalytic roles?
(1) 5 S rRNA
(2) 23 S rRNA
(3) 16 S rRNA
(4) 5.8 S rRNA
Explanation: 23 S rRNA serves as a scaffold for ribosomal proteins and catalyzes peptide bond formation, making it both structural and catalytic. Correct answer: (2) 23 S rRNA.
7. Assertion-Reason MCQ:
Assertion (A): 23 S rRNA acts as a ribozyme.
Reason (R): It catalyzes peptide bond formation in ribosomes.
(1) Both A and R are true and R explains A
(2) Both A and R are true but R does not explain A
(3) A is true but R is false
(4) A is false but R is true
Explanation: 23 S rRNA catalyzes peptide bond formation, proving its ribozyme function. Both assertion and reason are true, and the reason explains the assertion. Correct answer: (1) Both A and R are true and R explains A.
8. Matching Type MCQ:
Match rRNAs with their function:
A. 16 S rRNA
B. 23 S rRNA
C. 5 S rRNA
D. 5.8 S rRNA
1. Structural and catalytic in bacteria
2. Decoding mRNA in bacteria
3. Structural in large subunit
4. Component of eukaryotic large subunit
Options:
(1) A-2, B-1, C-3, D-4
(2) A-1, B-2, C-4, D-3
(3) A-3, B-4, C-1, D-2
(4) A-4, B-3, C-2, D-1
Explanation: Correct matches: 16 S rRNA-decoding mRNA (A-2), 23 S rRNA-structural and catalytic (B-1), 5 S rRNA-structural (C-3), 5.8 S rRNA-eukaryotic large subunit (D-4). Correct answer: (1) A-2, B-1, C-3, D-4.
9. Fill in the Blanks MCQ:
The bacterial rRNA that functions as a ribozyme is ______.
(1) 16 S rRNA
(2) 23 S rRNA
(3) 5 S rRNA
(4) 18 S rRNA
Explanation: 23 S rRNA of the large subunit acts as a ribozyme, catalyzing peptide bond formation while providing structural support. Correct answer: (2) 23 S rRNA.
10. Choose the Correct Statements MCQ:
Select correct statements:
(1) 23 S rRNA has catalytic activity
(2) 16 S rRNA participates in decoding
(3) 5 S rRNA catalyzes peptide bonds
(4) 5.8 S rRNA is bacterial
Options:
(1) 1 and 2 only
(2) 1, 2, 3 only
(3) 2 and 4 only
(4) All are correct
Explanation: Statements 1 and 2 are correct. 23 S rRNA has catalytic activity, and 16 S rRNA is involved in decoding mRNA. 5 S rRNA is structural but not catalytic, and 5.8 S rRNA is eukaryotic. Correct answer: (1) 1 and 2 only.