Biochemistry Topic: Cell Objectives: Describe the basic structure of human cell. Explain the function of human cell. Discuss components of the cell sub-cellular organelles and their structure in relation to each organelle Topic: Call Membrane Objectives Describe the structure of the cell membrane. Discuss the basic structure (histological and biochemical) of the cell membrane. Explain the function of the membrane proteins and the importance of the lipid bilayer Topic: Membrane transport system Objectives: Explain the term semi-permeable. Describe the passive transport mechanism across a biological membrane Explain active transport. Describe the role of ATP in active transport mechanism. Compare the mechanism of passive transport & active transport. Topic: Concept of pH and pH Scale Objectives Define "pH" and its importance. Explain Henderson-Hasselbalch equation and its application. Discuss different concept of acid & base. Discuss the pH value of important biological fluids & their importance. List the different methods for pH determination Topic: Buffers and their Mechanism of action Objectives: Define buffer Explain different types of physiological buffers and the important biological functions of these buffers. Discuss acidosis and alkalosis. Topic: Carbohydrates Objective: Define Carbohydrates. Explain the biochemical functions of carbohydrates. Describe the importance of dietary carbohydrates. Classify carbohydrates. Topic: Monosaccharaides-1 Objective: Describe the structure and functions of important monosaccharaides and their derivatives. Topic: Monosaccharaides-II elation Dilayer ons of Objective: Discuss the properties of monosaccharaides i.e. chiral carbon, Stereoisomers, empire, optical isomers glycosides, amino sugar, and deoxy sugar. Topic: Disaccharides Objective: Discuss the structure of different disaccharides. Describe biomedical importance of disaccharides Topic: Polysaccharides-1 Objective: Discuss structure and function of homo-polysaccharide. Topic: Polysaccharides-II Objective: Discuss structure and function of heteropolysaccharide. Discuss different clinical disorders related to carbohydrate. Topic: Protein-l Objective: Define amino acids. Describe the structure of amino acids. Explain the properties of amino acids Define zwitter ions & isoelectric pH Topic: Protein-II Objective: List the name of amino acids commonly occurs in proteins Classify amino acids on the basis of structures, nutritional & polarity. Discuss important function of amino acids Topic: Protein-III Objective: Define proteins Differentiate between good quality proteins & poor quality protein on the basis of characteristics and give examples. Topic Protein -IV Discuss the biomedical importance of protein. Classify protein on the basis of physicochemical properties & molecular length and shape Topic: Structure of Protein-l Objective: Discuss the primary and secondary structure of protein. Topic: Structure of Protein-II Objective: Discuss the tertiary & quaternary structure of protein. Topic: Lipids-1 Objective: Define Lipids. Classify lipids and their biochemical function. Classify fatty acids. Describe the structure of fatty acids and their biomedical functions. Topics: Lipids-11 Objectives: Define glycerol Discuss structure & function of glycerol. Define cholesterol. Discuss structure, biomedical importance, and normal blood levels of cholesterol Discuss occurrence & forms of cholesterol Topics: Lipids-III Objectives: Classify simple lipids Describe hydrolysis, saponification, iodine number Describe the lipid peroxidation Topics: Lipids-IV Objectives: Describe the structure & functions of phospholipids Discuss the structure function of glycolipids. Topics: Lipids-V Objectives: Explain eicosanoids production & their importance in health and disease. Topic: Enzymes-I Objective: Define enzymes. Describe Classification of enzymes. Explain the properties and functions of enzymes, co-enzymes and Co-factors. Topic: Enzymes-II Objective: Explain the lock & key and induced fit theory. Explain different factors affects enzyme activity Explain significance of km & Vmax Topic: Nucleic Acid-1 Objectives: Define nucleic acids Describe the types and functions of nucleic Acid Discuss the structure of nucleic acid. Comparison between DNA & RNA Topic: Nucleic Acid-II Objectives Define replication, transcription and translation Describe the structure of nucleoside & nucleotides and polynucleotide. Topic: Nucleic Acid-III Objectives Explain the nucleotide derivatives Discuss the biomedical importance of ATP and CAMP holesterol Topic: Genetic-l Learning Objectives: Define replication Justify different theories of replication. List the substrates involved in DNA replication. Explain the processes needed to facilitate replication Topic: Genetics-II Learning Objectives: Define transcription. Discuss the different functions of RNA polymerase. Name the inhibitors of transcription. Explain the steps involved in transcription. Topic: Genetics-III Learning Objectives: Define genetic code. Discuss the characteristics of genetic code Define translation. Enlist the requirements of translation, make hard mcqs for my module

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general-purpose

Generate 5 hard MCQs for each of the following 13 biochemistry topics. Each MCQ must have: a stem, 5 options (A–E), the correct answer, and a brief explanation. Make questions genuinely difficult — test application, mechanisms, clinical correlation, and exception knowledge rather than simple recall. Use formal biochemistry language. TOPICS: 1. Cell (basic structure & function of human cell, subcellular organelles — structure and function) 2. Cell Membrane (histological and biochemical structure, fluid mosaic model, lipid bilayer) 3. Membrane Transport System (semi-permeability, passive transport, active transport, ATP role, comparison) 4. Concept of pH and pH Scale (Henderson-Hasselbalch equation, acids/bases, biological fluid pH values, pH determination methods) 5. Buffers and their Mechanism of Action (buffer definition, physiological buffers, acidosis, alkalosis) 6. Carbohydrates (definition, biochemical functions, dietary importance, classification) 7. Monosaccharides-I (structure and functions of important monosaccharides and their derivatives) 8. Monosaccharides-II (chiral carbon, stereoisomers, epimers, optical isomers, glycosides, amino sugars, deoxy sugars) 9. Disaccharides (structure of different disaccharides, biomedical importance) 10. Polysaccharides-I (structure and function of homopolysaccharides) 11. Polysaccharides-II (structure and function of heteropolysaccharides, clinical disorders related to carbohydrates) 12. Protein-I (amino acid definition, structure, properties, zwitterions, isoelectric pH) 13. Protein-II (amino acids in proteins, classification by structure/nutritional/polarity, functions of amino acids) Format: --- TOPIC X: [Topic Name] --- Q1. [Stem] A. ... B. ... C. ... D. ... E. ... Answer: [Letter] Explanation: [1–2 sentences] Q2. ... (repeat for Q3, Q4, Q5) Return all 65 MCQs in plain text.

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general-purpose

Generate 5 hard MCQs for each of the following 13 biochemistry topics. Each MCQ must have: a stem, 5 options (A–E), the correct answer, and a brief explanation. Make questions genuinely difficult — test application, mechanisms, clinical correlation, and exception knowledge rather than simple recall. Use formal biochemistry language. TOPICS: 14. Protein-III (definition of proteins, good quality vs poor quality proteins — characteristics and examples) 15. Protein-IV (biomedical importance of proteins, classification by physicochemical properties, molecular length and shape) 16. Structure of Protein-I (primary and secondary structure of protein — alpha helix, beta sheet, etc.) 17. Structure of Protein-II (tertiary and quaternary structure of protein) 18. Lipids-I (definition, classification of lipids, biochemical functions, classification and structure of fatty acids, biomedical functions) 19. Lipids-II (glycerol structure/function, cholesterol structure/biomedical importance/normal blood levels/occurrence/forms) 20. Lipids-III (classification of simple lipids, hydrolysis, saponification, iodine number, lipid peroxidation) 21. Lipids-IV (structure and functions of phospholipids, structure and function of glycolipids) 22. Lipids-V (eicosanoids production, importance in health and disease) 23. Enzymes-I (definition, classification, properties and functions of enzymes, coenzymes, cofactors) 24. Enzymes-II (lock & key vs induced fit theory, factors affecting enzyme activity, Km and Vmax significance) 25. Nucleic Acid-I (definition, types and functions of nucleic acids, structure, comparison between DNA & RNA) 26. Nucleic Acid-II (replication, transcription, translation definitions; structure of nucleoside, nucleotides, polynucleotides) 27. Nucleic Acid-III (nucleotide derivatives, biomedical importance of ATP and cAMP) 28. Genetics-I (DNA replication — theories, substrates, processes/enzymes needed) 29. Genetics-II (transcription — RNA polymerase functions, inhibitors of transcription, steps of transcription) 30. Genetics-III (genetic code — definition and characteristics; translation — definition, requirements) Format: --- TOPIC X: [Topic Name] --- Q1. [Stem] A. ... B. ... C. ... D. ... E. ... Answer: [Letter] Explanation: [1–2 sentences] Q2. ... (repeat for Q3, Q4, Q5) Return all 65 MCQs (topics 14–26 plus Nucleic Acids and Genetics, listed as Topics 14-30 above) in plain text.

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okie i said mcqs like a mock paper

Here are all 130 hard MCQs across all 26 topics, formatted as a mock exam paper.

BIOCHEMISTRY MOCK PAPER

Cell Biology, Membrane Transport, pH & Buffers, Carbohydrates, Proteins, Lipids, Enzymes, Nucleic Acids & Genetics

Total Questions: 130 | 5 per Topic | Time: 2 hrs 10 min

SECTION A: CELL

Q1. Which organelle is responsible for post-translational modification, sorting, and packaging of glycoproteins for secretion?

A. Smooth endoplasmic reticulum
B. Rough endoplasmic reticulum
C. Golgi apparatus
D. Lysosome
E. Peroxisome

Answer: C The Golgi apparatus receives proteins from the RER, modifies them (glycosylation, phosphorylation), and packages them into vesicles for secretion or delivery to lysosomes.

Q2. A cell with a defective mitochondrial F₀F₁-ATPase will most likely show which of the following?

A. Impaired fatty acid synthesis
B. Impaired ATP synthesis from the electron transport chain
C. Impaired protein synthesis
D. Impaired lysosomal digestion
E. Impaired nuclear division

Answer: B F₀F₁-ATPase (ATP synthase) uses the proton gradient across the inner mitochondrial membrane to synthesize ATP; its defect uncouples oxidative phosphorylation.

Q3. The nuclear envelope is continuous with which cellular structure?

A. Plasma membrane
B. Golgi cisternae
C. Rough endoplasmic reticulum
D. Smooth endoplasmic reticulum
E. Lysosomal membrane

Answer: C The outer nuclear membrane is structurally continuous with the rough ER, allowing direct communication between the nuclear envelope and the ER lumen.

Q4. Peroxisomes are unique among cytoplasmic organelles because they:

A. Contain ribosomes on their membrane
B. Synthesize their own DNA
C. Generate H₂O₂ and contain catalase to degrade it
D. Are derived from the Golgi apparatus
E. Contain the enzymes of the citric acid cycle

Answer: C Peroxisomes oxidize long-chain fatty acids and amino acids producing H₂O₂, which is immediately detoxified by catalase — this is their defining biochemical feature.

Q5. Which of the following best describes the role of the smooth endoplasmic reticulum in hepatocytes?

A. Synthesis of secretory proteins
B. Drug detoxification via cytochrome P450 enzymes
C. ATP production via oxidative phosphorylation
D. Autophagy of damaged organelles
E. Ribosome assembly

Answer: B In hepatocytes, the smooth ER is rich in cytochrome P450 enzymes responsible for hydroxylation and detoxification of drugs, steroids, and xenobiotics.

SECTION B: CELL MEMBRANE

Q6. According to the fluid mosaic model, which property of membrane phospholipids is primarily responsible for the lateral fluidity of the membrane?

A. Length of the fatty acid chain
B. Degree of unsaturation of fatty acid tails
C. Presence of phosphate head groups
D. Abundance of integral proteins
E. Sphingomyelin content

Answer: B Unsaturated fatty acids introduce kinks that prevent tight packing of phospholipid tails, increasing membrane fluidity; cholesterol modulates this effect.

Q7. Which of the following best describes integral membrane proteins?

A. They are loosely attached to the outer leaflet only
B. They span the lipid bilayer and require detergents for extraction
C. They are anchored via GPI linkage exclusively
D. They associate only with peripheral proteins
E. They are found only on the intracellular face

Answer: B Integral (transmembrane) proteins span the bilayer via hydrophobic alpha-helical domains and require detergent treatment (or organic solvents) to be extracted.

Q8. The asymmetry of the plasma membrane lipid bilayer is maintained by which enzyme?

A. Phospholipase A₂
B. Flippase (aminophospholipid translocase)
C. Sphingomyelinase
D. Phospholipase C
E. Lysophospholipase

Answer: B Flippase actively moves phosphatidylserine and phosphatidylethanolamine from the outer to the inner leaflet using ATP, maintaining bilayer asymmetry.

Q9. Cholesterol increases membrane microviscosity at high temperatures but decreases it at low temperatures. This is because cholesterol:

A. Covalently crosslinks phospholipid head groups
B. Inserts between fatty acid tails, preventing both extreme packing and extreme fluidity
C. Activates membrane-bound Na⁺/K⁺-ATPase
D. Increases the proportion of saturated fatty acids
E. Reduces the number of transmembrane proteins

Answer: B Cholesterol's rigid sterol ring intercalates between phospholipid tails, acting as a "fluidity buffer" — it prevents excessive rigidity at low T and excessive fluidity at high T.

Q10. A mutation abolishing glycocalyx formation would most directly impair:

A. ATP synthesis at the inner membrane
B. Cell-cell recognition and receptor-ligand interactions
C. Active transport of glucose
D. Ion channel gating
E. Synthesis of transmembrane proteins

Answer: B The glycocalyx (oligosaccharide chains on glycoproteins/glycolipids) mediates cell-cell recognition, immune interactions, and receptor binding at the cell surface.

SECTION C: MEMBRANE TRANSPORT SYSTEM

Q11. Ouabain inhibits the Na⁺/K⁺-ATPase by competing with which ion at its binding site?

A. Na⁺
B. K⁺
C. Ca²⁺
D. Mg²⁺
E. H⁺

Answer: B Ouabain (a cardiac glycoside) binds the extracellular K⁺-binding site of the pump, blocking dephosphorylation and halting the transport cycle.

Q12. Glucose entry into erythrocytes occurs via GLUT1. This is best described as:

A. Primary active transport
B. Secondary active transport
C. Facilitated diffusion
D. Simple diffusion
E. Endocytosis

Answer: C GLUT1 transports glucose down its concentration gradient without energy expenditure; it is a carrier protein enabling facilitated (passive) diffusion.

Q13. In secondary active transport of glucose in intestinal epithelial cells, the energy driving glucose uptake is derived from:

A. Direct hydrolysis of ATP
B. The Na⁺ electrochemical gradient generated by Na⁺/K⁺-ATPase
C. The glucose concentration gradient alone
D. GTP hydrolysis
E. Membrane potential only

Answer: B SGLT1 co-transports Na⁺ and glucose; Na⁺ moves down its gradient (created by Na⁺/K⁺-ATPase) and drives glucose uphill — this is secondary active transport.

Q14. The term "semi-permeable" in the context of biological membranes best refers to:

A. Equal permeability to all solutes
B. Selective permeability based on solute size, charge, and lipid solubility
C. Permeability only to water molecules
D. Impermeability to all ions
E. Permeability dependent solely on temperature

Answer: B A semi-permeable membrane selectively allows passage based on physicochemical properties — small nonpolar molecules pass freely, while large or charged molecules require carriers or channels.

Q15. A cell placed in a hypotonic solution will undergo which of the following?

A. Crenation due to water efflux
B. Lysis due to water influx by osmosis
C. No change in volume
D. Increased Na⁺/K⁺-ATPase activity causing shrinkage
E. Facilitated glucose uptake

Answer: B In a hypotonic environment, water moves into the cell by osmosis (down its water potential gradient), causing the cell to swell and potentially lyse.

SECTION D: pH AND pH SCALE

Q16. Using the Henderson-Hasselbalch equation, calculate the pH of a bicarbonate buffer where [HCO₃⁻] = 24 mEq/L and [H₂CO₃] = 1.2 mEq/L (pKa = 6.1):

A. 6.1
B. 6.8
C. 7.0
D. 7.4
E. 7.8

Answer: D pH = 6.1 + log(24/1.2) = 6.1 + log(20) = 6.1 + 1.3 = 7.4; this represents normal arterial blood pH.

Q17. According to the Lewis acid-base concept, a Lewis acid is defined as:

A. A proton donor
B. A proton acceptor
C. An electron pair donor
D. An electron pair acceptor
E. A hydroxyl ion donor

Answer: D Lewis acids accept electron pairs (electrophiles), while Lewis bases donate electron pairs — this is broader than the Brønsted-Lowry definition.

Q18. The pH of cerebrospinal fluid (CSF) is approximately:

A. 6.8
B. 7.0
C. 7.3
D. 7.35
E. 7.8

Answer: C — 7.3 CSF has a pH of approximately 7.3–7.35, slightly lower than arterial blood (7.4), which is important for central chemoreceptor regulation of respiration.

Q19. A Brønsted-Lowry base differs from an Arrhenius base in that a Brønsted-Lowry base:

A. Must dissociate in water to release OH⁻
B. Can accept protons in non-aqueous solvents
C. Only functions in acidic solutions
D. Must contain nitrogen
E. Requires water as solvent

Answer: B Brønsted-Lowry bases are proton acceptors and can function in any solvent, whereas Arrhenius bases must produce OH⁻ in aqueous solution.

Q20. Which method of pH determination is most accurate for measuring blood pH in clinical settings?

A. Litmus paper
B. pH indicator dyes
C. Glass electrode pH meter
D. Universal indicator solution
E. Colorimetric strip testing

Answer: C The glass electrode pH meter measures the potential difference across a pH-sensitive glass membrane, providing the most accurate and reproducible clinical pH measurement.

SECTION E: BUFFERS AND MECHANISM OF ACTION

Q21. A patient with diabetic ketoacidosis has pH 7.1, HCO₃⁻ 8 mEq/L, PaCO₂ 20 mmHg. What is the primary acid-base disorder?

A. Respiratory acidosis
B. Metabolic alkalosis
C. Metabolic acidosis with respiratory compensation
D. Respiratory alkalosis
E. Mixed respiratory and metabolic alkalosis

Answer: C Low pH + low HCO₃⁻ = metabolic acidosis; the low PaCO₂ (normal ~40) reflects compensatory hyperventilation (Kussmaul breathing) to blow off CO₂.

Q22. The phosphate buffer system (H₂PO₄⁻/HPO₄²⁻) is most effective in which biological compartment?

A. Arterial blood plasma
B. Interstitial fluid
C. Intracellular fluid and urine
D. Cerebrospinal fluid
E. Lymph

Answer: C The phosphate buffer (pKa 6.8) is the major intracellular buffer and is especially important in renal tubular fluid (urine), where it excretes titratable acid.

Q23. Which of the following statements about the bicarbonate buffer system is correct?

A. It has a pKa of 7.4, making it ideal at physiological pH
B. It is effective despite its pKa (6.1) being away from physiological pH because CO₂ is an open system regulated by the lungs
C. It operates independently of renal regulation
D. It is the primary intracellular buffer
E. Carbonic anhydrase plays no role in this system

Answer: B Although pKa 6.1 is not ideal, the bicarbonate system is the most important blood buffer because lungs can rapidly adjust PaCO₂, making it an open system with enormous buffering capacity.

Q24. In respiratory alkalosis, renal compensation involves:

A. Increased H⁺ secretion into tubular lumen
B. Decreased reabsorption of HCO₃⁻ and increased excretion
C. Increased ammonia production
D. Retention of CO₂
E. Increased chloride reabsorption

Answer: B In respiratory alkalosis (low PaCO₂, high pH), the kidneys compensate by reducing HCO₃⁻ reabsorption and increasing its urinary excretion to lower plasma pH.

Q25. A buffer is most effective within how many pH units of its pKa?

A. ± 0.5
B. ± 1.0
C. ± 1.5
D. ± 2.0
E. ± 2.5

Answer: B A buffer resists pH change effectively within ±1 pH unit of its pKa, where the ratio of conjugate base to acid is between 10:1 and 1:10.

SECTION F: CARBOHYDRATES

Q26. Which of the following carbohydrates serves as the primary storage form of glucose in animal tissues?

A. Starch
B. Cellulose
C. Glycogen
D. Inulin
E. Dextrin

Answer: C Glycogen is the storage polysaccharide in animals (liver and muscle); starch serves this role in plants. Liver glycogen maintains blood glucose; muscle glycogen fuels local activity.

Q27. The general molecular formula for carbohydrates (Cₙ(H₂O)ₙ) is least applicable to which of the following?

A. Glucose
B. Fructose
C. Deoxyribose
D. Galactose
E. Ribose

Answer: C Deoxyribose (C₅H₁₀O₄) lacks one oxygen atom compared to the formula Cₙ(H₂O)ₙ — it is a deoxy sugar and therefore does not strictly fit the carbohydrate empirical formula.

Q28. Dietary fiber (non-starch polysaccharides) primarily provides health benefit by:

A. Being directly absorbed in the small intestine
B. Providing 9 kcal/g of energy
C. Fermentation by colonic bacteria producing short-chain fatty acids and reducing transit time
D. Stimulating pancreatic amylase secretion
E. Being hydrolyzed to glucose by intestinal glucosidases

Answer: C Dietary fiber is fermented by colonic microbiota into short-chain fatty acids (butyrate, propionate, acetate) and reduces bowel transit time, lowering colorectal cancer risk.

Q29. Which carbon of glucose is the reference carbon in determining D- versus L-configuration?

A. C-1
B. C-2
C. C-3
D. C-5
E. C-6

Answer: D In a monosaccharide, the D/L designation is based on the configuration of the highest-numbered chiral carbon — C-5 in hexoses (like glucose), compared to D-glyceraldehyde.

Q30. Galactosemia results from a deficiency of which enzyme?

A. Glucose-6-phosphatase
B. Galactose-1-phosphate uridylyl transferase
C. Fructokinase
D. Aldolase B
E. Phosphoglucomutase

Answer: B Classic galactosemia is caused by deficiency of galactose-1-phosphate uridylyl transferase, causing accumulation of galactose-1-phosphate toxic to liver, brain, and kidneys.

SECTION G: MONOSACCHARIDES-I

Q31. Which monosaccharide derivative is an important component of the anticoagulant heparin?

A. Glucuronic acid
B. Galactose
C. Fructose-6-phosphate
D. Mannose
E. Xylose

Answer: A Heparin is a glycosaminoglycan containing alternating units of iduronic acid and glucosamine sulfate (plus glucuronic acid), giving it its strong negative charge and anticoagulant activity.

Q32. N-acetylneuraminic acid (sialic acid) is biochemically derived from which two precursors?

A. Glucose + Glutamine
B. Mannose-6-phosphate + Pyruvate
C. UDP-N-acetylmannosamine + Phosphoenolpyruvate
D. Fructose-6-phosphate + Acetyl-CoA
E. Galactose + UDP-glucuronic acid

Answer: C Sialic acid (NANA) is synthesized from UDP-N-acetylmannosamine and phosphoenolpyruvate; it is found on cell surface glycoproteins and glycolipids, playing roles in cell recognition.

Q33. Glucose-6-phosphate plays a pivotal role in which of the following metabolic pathways? (Select the MOST comprehensive answer)

A. Glycolysis only
B. Glycolysis and glycogenesis only
C. Glycolysis, glycogenesis, pentose phosphate pathway, and gluconeogenesis
D. Gluconeogenesis and urea cycle
E. Beta-oxidation and ketogenesis

Answer: C G-6-P is a branch-point metabolite feeding into glycolysis, glycogen synthesis, the pentose phosphate pathway, and (via glucose-6-phosphatase in liver) gluconeogenesis output.

Q34. Glucuronic acid is important in the body primarily because it:

A. Serves as the main energy source for erythrocytes
B. Is a component of glycosaminoglycans and conjugates toxins/bilirubin for excretion
C. Is the precursor for ketone body synthesis
D. Provides ribose for nucleotide synthesis
E. Is the primary substrate for glycogen synthesis

Answer: B Glucuronic acid is found in hyaluronic acid, heparin, and chondroitin sulfate; in the liver it conjugates bilirubin, drugs, and steroids (glucuronidation) for urinary/biliary excretion.

Q35. Which monosaccharide is the preferred fuel for the brain under normal (non-fasting) conditions, and what transporter mediates its entry across the blood-brain barrier?

A. Fructose via GLUT5
B. Glucose via GLUT1
C. Galactose via GLUT2
D. Glucose via GLUT4
E. Mannose via SGLT1

Answer: B The brain preferentially uses glucose as fuel under fed conditions; GLUT1 (a high-affinity, insulin-independent transporter) facilitates its entry across the blood-brain barrier.

SECTION H: MONOSACCHARIDES-II

Q36. Glucose and galactose are epimers because they differ in configuration at:

A. C-1
B. C-2
C. C-4
D. C-5
E. C-6

Answer: C Epimers differ at only one chiral carbon. Glucose and galactose differ at C-4 (axial vs equatorial –OH), making them C-4 epimers.

Q37. The optical rotation of a mixture of equal amounts of D- and L-glucose would be:

A. +52.7° (dextrorotatory)
B. -52.7° (levorotatory)
C. 0° (optically inactive — racemic mixture)
D. +93° (alpha anomer)
E. +19° (beta anomer)

Answer: C A racemic mixture contains equal amounts of enantiomers (D and L forms); their equal and opposite optical rotations cancel to give net zero optical activity.

Q38. N-acetylglucosamine (GlcNAc) is the monomer unit of:

A. Starch
B. Cellulose
C. Hyaluronic acid and chitin
D. Heparan sulfate only
E. Dermatan sulfate only

Answer: C GlcNAc is the building block of chitin (β-1,4 linkages) and alternates with glucuronic acid in hyaluronic acid; it is also found in other glycosaminoglycans.

Q39. In the formation of an O-glycoside, the glycosidic bond forms between:

A. C-1 of the sugar and a nitrogen atom of an amino acid
B. C-1 of the sugar and a hydroxyl group of another molecule
C. C-2 of the sugar and a carboxyl group
D. C-6 of the sugar and a phosphate group
E. Two anomeric carbons of different sugars

Answer: B O-glycosides form when the anomeric C-1 reacts with a hydroxyl group (-OH) of an alcohol or another sugar; N-glycosides form with amino groups.

Q40. A compound with 4 chiral carbons has how many possible stereoisomers?

A. 4
B. 8
C. 16
D. 12
E. 6

Answer: C The maximum number of stereoisomers = 2ⁿ, where n = number of chiral carbons. 2⁴ = 16 possible stereoisomers.

SECTION I: DISACCHARIDES

Q41. Lactose intolerance results from deficiency of lactase. Which products normally result from lactase action on lactose?

A. Glucose + Fructose
B. Glucose + Galactose
C. Two molecules of Glucose
D. Galactose + Fructose
E. Glucose + Mannose

Answer: B Lactose (β-galactosidase substrate) is a β-1,4-glycosidic bond between galactose (C-1) and glucose (C-4); hydrolysis yields glucose + galactose.

Q42. Sucrose is a non-reducing sugar because:

A. It contains only fructose residues
B. Both anomeric carbons (C-1 of glucose and C-2 of fructose) are involved in the glycosidic bond
C. It lacks a hydroxyl group at C-4
D. It is an α-1,6-linked disaccharide
E. It contains an N-glycosidic bond

Answer: B Sucrose has an α,β-1,2 glycosidic bond linking C-1 of glucose to C-2 of fructose, locking both anomeric carbons and eliminating the free reducing group.

Q43. Trehalose (α-1,1-glycosidic bond) is found in which organism and serves what primary function?

A. Human muscle — energy storage
B. Insects and fungi — protection against desiccation and freezing
C. Plants — structural support
D. Bacteria — cell wall integrity
E. Humans — blood group determinants

Answer: B Trehalose is abundant in insects, fungi, and yeast; it acts as an anhydrobiosis protectant, replacing water around membrane phospholipids during desiccation or cold stress.

Q44. Maltose is produced during starch digestion by salivary and pancreatic amylase. Its glycosidic linkage is:

A. β-1,4-glucose-glucose
B. α-1,6-glucose-glucose
C. α-1,4-glucose-glucose
D. β-1,6-glucose-galactose
E. α-1,2-glucose-fructose

Answer: C Maltose consists of two glucose units joined by an α-1,4-glycosidic bond, produced when amylase cleaves the α-1,4 linkages of starch.

Q45. A neonate presents with jaundice, cataracts, and E. coli sepsis in the first week of life. Which disaccharide is most relevant to this clinical picture?

A. Sucrose
B. Maltose
C. Lactose
D. Trehalose
E. Cellobiose

Answer: C This is classic galactosemia from lactose ingestion (breast milk); galactose-1-phosphate accumulates causing liver damage (jaundice), cataracts, and susceptibility to E. coli sepsis.

SECTION J: POLYSACCHARIDES-I (HOMOPOLYSACCHARIDES)

Q46. Glycogen differs structurally from amylopectin in that glycogen:

A. Contains only α-1,4 linkages
B. Has α-1,6 branch points every 8–12 glucose units (vs every 24–30 in amylopectin)
C. Is composed of β-1,4-linked glucose
D. Does not have a reducing end
E. Is found only in hepatocytes

Answer: B Glycogen is more highly branched than amylopectin — branching every 8–12 residues in glycogen vs every 24–30 in amylopectin — allowing faster simultaneous phosphorolysis.

Q47. Cellulose cannot be digested by humans because we lack:

A. Alpha-amylase acting on β-1,4 bonds
B. Beta-glucosidase (cellulase) acting on β-1,4-glycosidic bonds
C. Maltase acting on α-1,4 bonds
D. Isomaltase acting on α-1,6 bonds
E. Sucrase acting on α-1,2 bonds

Answer: B Cellulose is a β-1,4-linked glucose polymer; humans lack cellulase (β-1,4-glucosidase). Ruminants can digest it via microbial cellulase in their rumen.

Q48. The enzyme responsible for creating branch points during glycogen synthesis is:

A. Glycogen phosphorylase
B. Glycogen synthase
C. Branching enzyme (α-1,4 to α-1,6 glucan transferase)
D. Debranching enzyme
E. Phosphoglucomutase

Answer: C Branching enzyme transfers a terminal 6–7 residue chain from an α-1,4 linkage to create a new α-1,6 branch point during glycogen synthesis.

Q49. In Von Gierke disease (Type I glycogen storage disease), the enzyme deficient is:

A. Lysosomal α-1,4-glucosidase
B. Glucose-6-phosphatase
C. Liver phosphorylase
D. Debranching enzyme
E. Glycogen synthase

Answer: B Von Gierke disease is due to glucose-6-phosphatase deficiency, causing massive hepatomegaly, severe fasting hypoglycemia, hyperlipidemia, and lactic acidosis.

Q50. Inulin (a fructan) is used clinically to measure:

A. Creatinine clearance
B. Glomerular filtration rate (GFR)
C. Tubular reabsorption capacity
D. Hepatic blood flow
E. Intestinal permeability

Answer: B Inulin is freely filtered at the glomerulus and neither secreted nor reabsorbed by renal tubules — making it the gold standard for measuring GFR.

SECTION K: POLYSACCHARIDES-II (HETEROPOLYSACCHARIDES & CLINICAL DISORDERS)

Q51. Hyaluronic acid is unique among glycosaminoglycans because it:

A. Contains sulfate groups
B. Is covalently linked to a core protein
C. Is not sulfated and is not linked to a core protein (it is a free GAG)
D. Contains only N-acetylgalactosamine
E. Is found exclusively in cartilage

Answer: C Hyaluronic acid is non-sulfated and exists as a free polysaccharide (not attached to a core protein), unlike other GAGs which form proteoglycans.

Q52. In Hurler syndrome (Mucopolysaccharidosis Type I), the enzyme deficiency leads to accumulation of which substrates?

A. Keratan sulfate and heparan sulfate
B. Dermatan sulfate and heparan sulfate
C. Chondroitin-6-sulfate only
D. Hyaluronic acid and chondroitin sulfate
E. Heparin and heparan sulfate

Answer: B Hurler syndrome results from α-L-iduronidase deficiency, leading to accumulation of dermatan sulfate and heparan sulfate in lysosomes, causing coarse facies, organomegaly, and corneal clouding.

Q53. Blood group antigens (A, B, H) are determined by oligosaccharides attached to:

A. Plasma fibrinogen
B. Glycolipids and glycoproteins on erythrocyte surfaces
C. Intracellular glycogen
D. Circulating albumin
E. Nuclear histones

Answer: B ABO blood group antigens are oligosaccharide chains on glycolipids (glycosphingolipids) and glycoproteins displayed on the red blood cell surface.

Q54. Heparin's anticoagulant mechanism involves:

A. Direct inhibition of thrombin only
B. Activation of antithrombin III, which then inhibits thrombin and Factor Xa
C. Blocking vitamin K-dependent clotting factor synthesis
D. Dissolving existing fibrin clots
E. Inhibiting platelet COX-1

Answer: B Heparin binds antithrombin III, causing a conformational change that dramatically accelerates its inhibition of thrombin (IIa) and factor Xa, preventing fibrin clot formation.

Q55. Essential fructosuria differs from hereditary fructose intolerance in that:

A. Both are asymptomatic
B. Essential fructosuria is benign (fructokinase deficiency) while hereditary fructose intolerance (aldolase B deficiency) causes severe hypoglycemia and liver damage
C. Both involve aldolase B deficiency
D. Essential fructosuria causes cataracts
E. Hereditary fructose intolerance is the milder condition

Answer: B Essential fructosuria lacks fructokinase — fructose cannot be phosphorylated and is excreted harmlessly. Hereditary fructose intolerance accumulates fructose-1-phosphate (toxic), causing liver failure and hypoglycemia.

SECTION L: PROTEIN-I

Q56. At the isoelectric point (pI) of an amino acid, the net charge is zero. For aspartic acid (pKa₁=2.0, pKa₂=3.9, pKa₃=9.9), the pI is:

A. (2.0 + 9.9)/2 = 5.95
B. (2.0 + 3.9)/2 = 2.95
C. (3.9 + 9.9)/2 = 6.9
D. 3.9
E. 9.9

Answer: B For acidic amino acids, pI = (pKa₁ + pKa₂)/2 = (pKa of alpha-COOH + pKa of R-group COOH) = (2.0 + 3.9)/2 = 2.95. The pI lies between the two acidic pKa values.

Q57. Which of the following amino acids is capable of forming a disulfide bond?

A. Serine
B. Threonine
C. Cysteine
D. Methionine
E. Tyrosine

Answer: C Cysteine contains a thiol (–SH) group that can be oxidized to form disulfide bonds (–S–S–) between two cysteine residues — critical for tertiary protein structure.

Q58. In a zwitterion form of an amino acid, which statement is correct?

A. Both amino and carboxyl groups are uncharged
B. The amino group is protonated (–NH₃⁺) and the carboxyl group is deprotonated (–COO⁻)
C. Only the R-group carries charge
D. The molecule carries a net positive charge
E. The molecule carries a net negative charge

Answer: B At physiological pH (near pI), the amino acid exists as a zwitterion with –NH₃⁺ (protonated) and –COO⁻ (deprotonated), resulting in zero net charge.

Q59. Which property of amino acids allows them to act as biological buffers?

A. Their hydrophobic R-groups
B. Their amphoteric nature — ability to act as both acids and bases due to –NH₃⁺/–NH₂ and –COOH/–COO⁻ equilibria
C. Their ability to form peptide bonds
D. Their solubility in organic solvents
E. Their optical rotation

Answer: B Amino acids are amphoteric — they have both acidic (–COOH) and basic (–NH₂) groups, allowing them to donate or accept protons and thus buffer both acid and base loads.

Q60. Proline is unique among the 20 standard amino acids because:

A. It is the only amino acid with an aromatic ring
B. Its side chain forms a covalent bond with its own alpha-amino group, creating a secondary (imino) amine
C. It is the only non-chiral amino acid
D. It contains a sulfur atom
E. It has the highest molecular weight

Answer: B Proline's R-group cyclizes back to bond the nitrogen of the alpha-amino group, creating a pyrrolidine ring — making it a secondary amine (imino acid) that introduces rigid kinks in polypeptide chains.

SECTION M: PROTEIN-II

Q61. Which amino acid is classified as conditionally essential (essential only in premature neonates and in certain disease states)?

A. Leucine
B. Methionine
C. Arginine
D. Tryptophan
E. Valine

Answer: C Arginine is synthesized via the urea cycle but in insufficient amounts during rapid growth (premature infants) or severe illness, making it conditionally essential in these states.

Q62. The classification of amino acids as "polar uncharged" includes which of the following?

A. Aspartate, glutamate, lysine
B. Serine, threonine, asparagine, glutamine
C. Valine, leucine, isoleucine, phenylalanine
D. Histidine, arginine, lysine
E. Cysteine, methionine, tryptophan

Answer: B Serine, threonine, asparagine, and glutamine contain polar groups (–OH, –CONH₂) that can form hydrogen bonds but carry no charge at physiological pH.

Q63. Tryptophan is the precursor for which two important biomolecules?

A. Epinephrine and dopamine
B. Serotonin (5-HT) and NAD⁺/NADP⁺ (via nicotinamide)
C. Histamine and GABA
D. Thyroid hormone and melanin
E. Heme and bile salts

Answer: B Tryptophan is converted to serotonin (via 5-hydroxytryptophan) and to niacin/NAD⁺ (via the kynurenine pathway); deficiency causes pellagra-like symptoms.

Q64. Which amino acid serves as the major carrier of nitrogen from peripheral tissues to the liver?

A. Glutamate
B. Aspartate
C. Alanine
D. Glycine
E. Serine

Answer: C In the glucose-alanine cycle, alanine carries amino groups from muscle to liver; pyruvate in muscle accepts the amino group (from transamination) and is transported to the liver as alanine.

Q65. Phenylketonuria (PKU) results from deficiency of phenylalanine hydroxylase, causing accumulation of phenylalanine. This damages the brain primarily by:

A. Inhibiting gluconeogenesis in neurons
B. Competitively inhibiting transport of other large neutral amino acids across the blood-brain barrier
C. Directly destroying myelin
D. Causing cerebral lactic acidosis
E. Depleting brain glycogen stores

Answer: B Excess phenylalanine competes with tyrosine, tryptophan, and other large neutral amino acids for the same BBB transporter (LAT1), starving the brain of essential amino acids needed for neurotransmitter synthesis.

SECTION N: PROTEIN-III

Q66. A "complete" (high biological value) protein is defined as one that:

A. Contains all 20 amino acids in equal proportions
B. Contains all essential amino acids in proportions adequate to support nitrogen equilibrium and growth
C. Has molecular weight >100 kDa
D. Is entirely composed of non-polar amino acids
E. Can be synthesized entirely by the human body

Answer: B A complete protein provides all 9 essential amino acids in sufficient quantities. Animal proteins (eggs, meat, dairy) are generally complete; most plant proteins are limiting in one or more EAAs.

Q67. The "limiting amino acid" in a dietary protein is:

A. The amino acid present in the highest concentration
B. The essential amino acid present in the lowest concentration relative to human requirements
C. Any non-essential amino acid
D. The amino acid with the highest molecular weight
E. The first amino acid in the peptide sequence

Answer: B The limiting amino acid determines the extent to which the protein can be used for synthesis; for example, lysine is the limiting amino acid in wheat and methionine in legumes.

Q68. Protein digestibility corrected amino acid score (PDCAAS) measures:

A. Total protein content per gram of food
B. Amino acid profile relative to human requirements, corrected for digestibility
C. Rate of gastric protein hydrolysis
D. Serum albumin levels after protein ingestion
E. Urea nitrogen excretion after protein intake

Answer: B PDCAAS = (amino acid score × digestibility); it is the FAO/WHO standard for measuring protein quality, with a maximum score of 1.0 (e.g., whey, eggs).

Q69. Gelatin is a poor-quality protein because it:

A. Contains too many essential amino acids
B. Lacks tryptophan and has very low levels of other essential amino acids
C. Is completely indigestible
D. Contains excessive methionine
E. Has a PDCAAS of 1.0

Answer: B Gelatin (derived from collagen) completely lacks tryptophan and is very low in cysteine, tyrosine, and other EAAs — it cannot independently sustain protein synthesis.

Q70. Nitrogen balance is NEGATIVE in which of the following states?

A. Pregnancy
B. Growing children
C. Recovery from illness
D. Severe burns or trauma
E. Athletes in training

Answer: D Severe burns/trauma causes massive protein catabolism exceeding intake, resulting in negative nitrogen balance (N excretion > N intake). The other states represent anabolic conditions with positive balance.

SECTION O: PROTEIN-IV

Q71. Which of the following proteins is classified as a globular protein?

A. Collagen
B. Keratin
C. Myosin
D. Hemoglobin
E. Fibrin

Answer: D Hemoglobin is a globular protein (compact, roughly spherical, soluble). Collagen, keratin, and fibrin are fibrous proteins with elongated, structural functions.

Q72. Plasma albumin's most important clinical function is:

A. Oxygen transport
B. Immune defense
C. Maintenance of colloid oncotic pressure and transport of poorly soluble molecules
D. Enzymatic coagulation
E. Hormone synthesis

Answer: C Albumin (MW ~67 kDa) accounts for ~80% of plasma colloid osmotic pressure; it also transports fatty acids, bilirubin, drugs, and hormones.

Q73. A patient with nephrotic syndrome loses large amounts of albumin in urine. The most immediate clinical consequence is:

A. Polycythemia
B. Pitting edema due to reduced plasma oncotic pressure
C. Hypercoagulability only
D. Hypertension
E. Jaundice

Answer: B Loss of albumin reduces plasma oncotic pressure, allowing fluid to shift from intravascular to interstitial space, causing generalized pitting edema.

Q74. Metalloproteins are classified based on their prosthetic group. Which of the following is a metalloprotein containing copper?

A. Hemoglobin
B. Ceruloplasmin
C. Ferritin
D. Myoglobin
E. Cytochrome c

Answer: B Ceruloplasmin is the major copper-carrying protein in plasma (contains 6–7 Cu²⁺ atoms) and acts as a ferroxidase. Deficiency causes Wilson's disease complications.

Q75. Which chromatographic technique separates proteins based on molecular size?

A. Ion exchange chromatography
B. Affinity chromatography
C. Gel filtration (size exclusion) chromatography
D. Electrophoresis
E. Isoelectric focusing

Answer: C Gel filtration (size exclusion) chromatography uses porous beads; smaller proteins enter pores and elute later, while larger proteins are excluded and elute first — separation is by molecular size.

SECTION P: STRUCTURE OF PROTEIN-I

Q76. The primary structure of a protein refers to:

A. The 3D folding pattern stabilized by hydrogen bonds
B. The linear sequence of amino acids linked by peptide bonds
C. The association of multiple polypeptide subunits
D. Coiling into alpha-helices
E. Formation of beta-pleated sheets

Answer: B Primary structure is the specific sequence of amino acids in a polypeptide chain, held together by covalent peptide bonds — this sequence determines all higher-order structures.

Q77. The alpha-helix is stabilized primarily by:

A. Disulfide bonds between cysteine residues
B. Hydrophobic interactions between R-groups
C. Hydrogen bonds between the –NH of one peptide bond and the –C=O of the peptide bond four residues ahead (i → i+4)
D. Ionic interactions between charged R-groups
E. Van der Waals forces between adjacent turns

Answer: C The alpha-helix is stabilized by intramolecular hydrogen bonds between the backbone NH and C=O groups of residues separated by 4 positions (i to i+4), giving 3.6 residues per turn.

Q78. Which amino acid cannot be incorporated into an alpha-helix because its R-group bonds back to the backbone nitrogen, introducing a rigid kink?

A. Glycine
B. Alanine
C. Proline
D. Valine
E. Leucine

Answer: C Proline's cyclic pyrrolidine ring prevents rotation around the N–Cα bond and lacks an NH for hydrogen bonding — it is a "helix breaker" that introduces bends or turns.

Q79. In a parallel beta-pleated sheet, the hydrogen bonds are:

A. Perpendicular to the strand direction and the strands run in the same N→C direction
B. Along the strand direction and strands run antiparallel
C. Between every fourth residue within the same strand
D. Formed only between cysteine residues
E. Between alpha-helices only

Answer: A In parallel beta-sheets, all strands run in the same N→C direction, with hydrogen bonds angled obliquely between strands. Antiparallel sheets have strands running in opposite directions with more favorable, direct H-bonds.

Q80. A missense mutation converting glutamate to valine at position 6 of the beta-globin chain causes sickle cell disease. The mechanism of sickling involves:

A. Disruption of primary structure preventing chain synthesis
B. Replacement of a charged hydrophilic residue with a nonpolar hydrophobic residue that promotes abnormal polymerization under deoxygenation
C. Formation of an extra disulfide bond
D. Loss of the heme-binding pocket
E. Alteration of the Bohr effect

Answer: B Val replaces Glu at β6; the hydrophobic valine fits into a complementary hydrophobic "pocket" on adjacent HbS molecules, causing deoxygenated HbS to polymerize into long fibers that distort RBCs into sickle shape.

SECTION Q: STRUCTURE OF PROTEIN-II

Q81. The tertiary structure of a protein is primarily stabilized by which type of interaction in the hydrophobic core?

A. Peptide bonds
B. Hydrophobic interactions between nonpolar R-groups
C. Disulfide bonds only
D. Ionic bonds only
E. Hydrogen bonds between backbone atoms only

Answer: B The major driving force for tertiary structure folding is the hydrophobic effect — nonpolar R-groups cluster in the protein interior away from water, with H-bonds, ionic bonds, and disulfide bonds providing additional stability.

Q82. Quaternary structure is found in which of the following proteins?

A. Myoglobin
B. Insulin (single chain form)
C. Hemoglobin
D. Lysozyme
E. Ribonuclease A

Answer: C Hemoglobin has quaternary structure — it consists of 4 polypeptide subunits (2α + 2β) assembled non-covalently. Myoglobin is monomeric (no quaternary structure).

Q83. Protein denaturation involves:

A. Hydrolysis of peptide bonds
B. Disruption of secondary, tertiary, and quaternary structures without breaking the primary structure
C. Permanent loss of amino acids
D. Covalent modification of the polypeptide backbone
E. Synthesis of new proteins

Answer: B Denaturation disrupts non-covalent interactions (H-bonds, hydrophobic, ionic) and disulfide bonds, unfolding the protein — but the peptide backbone (primary structure) remains intact.

Q84. Chaperone proteins (e.g., HSP70) assist in protein folding by:

A. Providing energy via ATP hydrolysis to prevent misfolding and aggregation of nascent polypeptides
B. Synthesizing the polypeptide chain
C. Adding post-translational modifications
D. Degrading incorrectly folded proteins
E. Forming permanent structural components of the final protein

Answer: A Molecular chaperones bind exposed hydrophobic segments of unfolded proteins using ATP hydrolysis, preventing premature or incorrect aggregation and allowing proper folding.

Q85. In prion diseases, the infectious agent PrPSc differs from the normal PrPc in:

A. Primary amino acid sequence
B. Presence of additional amino acids
C. Conformational change — shift from alpha-helix–rich to beta-sheet–rich tertiary structure
D. Loss of glycosylation
E. Absence of disulfide bonds

Answer: C Prions involve no genetic mutation — PrPSc has the same primary structure as PrPc but adopts a misfolded beta-sheet conformation, making it protease-resistant and capable of templating misfolding of normal protein.

SECTION R: LIPIDS-I

Q86. Which fatty acid is classified as an omega-3 polyunsaturated fatty acid?

A. Arachidonic acid (20:4, Δ5,8,11,14)
B. Oleic acid (18:1, Δ9)
C. Eicosapentaenoic acid — EPA (20:5, Δ5,8,11,14,17)
D. Palmitoleic acid (16:1, Δ9)
E. Linoleic acid (18:2, Δ9,12)

Answer: C EPA (20:5, ω-3) has its first double bond at the 3rd carbon from the methyl (omega) end — defining it as omega-3. Linoleic acid is ω-6; arachidonic acid is ω-6.

Q87. Trans fatty acids differ from cis fatty acids in that trans fatty acids:

A. Raise HDL and lower LDL cholesterol
B. Are found abundantly in fish oils
C. Have a more linear conformation, raise LDL, and lower HDL, increasing cardiovascular risk
D. Are always saturated
E. Cannot be incorporated into membrane phospholipids

Answer: C Trans unsaturated fats (from hydrogenation) adopt a straighter configuration similar to saturated fats. They raise LDL, lower HDL, and increase cardiovascular disease risk more than saturated fats per gram.

Q88. The essential fatty acid linoleic acid (18:2, ω-6) is "essential" because:

A. It cannot be elongated in the body
B. Humans lack Δ12 and Δ15 desaturases needed to introduce double bonds beyond Δ9
C. It is not absorbed from the diet
D. It is only found in animal fats
E. It cannot be oxidized by beta-oxidation

Answer: B Humans can introduce double bonds only up to Δ9 (using Δ9-desaturase) but lack Δ12 and Δ15 desaturases; thus linoleic (ω-6) and alpha-linolenic acid (ω-3) must come from the diet.

Q89. Which of the following lipids serves as a precursor for steroid hormones, bile acids, and vitamin D?

A. Triglycerides
B. Phosphatidylcholine
C. Cholesterol
D. Ceramide
E. Sphingomyelin

Answer: C Cholesterol is the biosynthetic precursor for all steroid hormones (cortisol, aldosterone, sex hormones), bile acids, and vitamin D₃; it is also a critical membrane component.

Q90. A fatty acid designated as 20:4 Δ5,8,11,14 is:

A. A saturated fatty acid with 20 carbons
B. Arachidonic acid — a polyunsaturated ω-6 fatty acid with 4 double bonds starting at C-5
C. An omega-3 fatty acid
D. A monounsaturated fatty acid
E. A medium-chain fatty acid

Answer: B 20:4 Δ5,8,11,14 is arachidonic acid — 20 carbons, 4 double bonds at C5, C8, C11, C14; it is an ω-6 PUFA and the precursor for prostaglandins, thromboxanes, and leukotrienes.

SECTION S: LIPIDS-II

Q91. The normal fasting serum total cholesterol level in adults should ideally be below:

A. 150 mg/dL
B. 200 mg/dL
C. 240 mg/dL
D. 300 mg/dL
E. 100 mg/dL

Answer: B Desirable total serum cholesterol is <200 mg/dL; borderline high is 200–239 mg/dL; high is ≥240 mg/dL, per ATP III/ACC-AHA cardiovascular risk guidelines.

Q92. Cholesterol is primarily synthesized in the body by which pathway, and where?

A. Beta-oxidation pathway, in mitochondria of all cells
B. Mevalonate (HMG-CoA reductase) pathway, primarily in the liver and intestine
C. Pentose phosphate pathway, in erythrocytes
D. Fatty acid synthesis pathway, in adipose tissue
E. Purine synthesis pathway, in all rapidly dividing cells

Answer: B Cholesterol synthesis proceeds via the mevalonate pathway: Acetyl-CoA → HMG-CoA → Mevalonate (rate-limiting step: HMG-CoA reductase, inhibited by statins) → → Cholesterol; primarily in liver.

Q93. Esterified cholesterol differs from free cholesterol in that esterified cholesterol:

A. Is more hydrophilic and found on cell surfaces
B. Has a fatty acid attached at the C-3 hydroxyl group by LCAT or ACAT, making it more hydrophobic
C. Is the predominant form in cell membranes
D. Cannot be transported in plasma lipoproteins
E. Is the precursor for steroid hormones

Answer: B Cholesterol ester has a fatty acid esterified to the C-3 –OH, making it highly nonpolar; it is stored in lipid droplets or transported in the hydrophobic core of lipoproteins (LDL, HDL).

Q94. A patient presents with xanthomas, corneal arcus, and LDL cholesterol of 350 mg/dL. This is consistent with which condition?

A. Tangier disease
B. Familial hypercholesterolemia (defective LDL receptor)
C. Abetalipoproteinemia
D. Lipoprotein lipase deficiency
E. Fish-eye disease

Answer: B Familial hypercholesterolemia is caused by LDL receptor mutations; markedly elevated LDL leads to cholesterol deposition in tendons (xanthomas), arteries (premature atherosclerosis), and cornea (arcus).

Q95. Glycerol is phosphorylated to glycerol-3-phosphate primarily by glycerol kinase, which is NOT expressed in:

A. Liver
B. Kidney
C. Adipose tissue
D. Intestine
E. Brain

Answer: C Adipose tissue lacks glycerol kinase; it must obtain glycerol-3-phosphate from glycolysis (via DHAP reduction) to re-esterify fatty acids — this explains why active glycolysis is needed for fat storage in adipocytes.

SECTION T: LIPIDS-III

Q96. The iodine number (iodine value) of a fat measures:

A. The degree of rancidity
B. The number of ester bonds available for saponification
C. The degree of unsaturation — grams of iodine absorbed per 100 g of fat
D. The fatty acid chain length
E. The melting point of the fat

Answer: C The iodine value measures grams of I₂ absorbed per 100 g of fat; double bonds react with iodine — a high iodine number indicates high unsaturation (e.g., fish oil ~170; butter ~35).

Q97. Lipid peroxidation of polyunsaturated fatty acids is initiated by:

A. Saponification with NaOH
B. Abstraction of a hydrogen atom from a bis-allylic carbon by a free radical, generating a lipid radical
C. Hydrolysis by phospholipase A₂
D. Beta-oxidation in mitochondria
E. Esterification with glycerol

Answer: B Lipid peroxidation begins when a reactive oxygen species (·OH, etc.) abstracts an H from a bis-allylic carbon of a PUFA, generating a carbon radical that reacts with O₂ to form peroxyl radicals, propagating chain reactions.

Q98. Saponification number is defined as:

A. Grams of iodine absorbed per 100g fat
B. Milligrams of KOH required to saponify 1 gram of fat
C. Number of double bonds in a fatty acid
D. Percentage of free fatty acids in a fat
E. Melting point in degrees Celsius

Answer: B The saponification number = mg KOH to completely saponify 1 g of fat. Higher saponification numbers indicate shorter-chain fatty acids (more ester bonds per gram); used to assess purity of fats/oils.

Q99. Rancidity in fats is caused by:

A. Increased degree of saturation
B. Oxidative rancidity (free radical attack on unsaturated FAs) and hydrolytic rancidity (esterase/moisture hydrolysis releasing short-chain fatty acids)
C. Increased iodine number
D. Saponification with alkali
E. Addition of antioxidants

Answer: B Oxidative rancidity: free radicals attack double bonds of PUFAs producing aldehydes/ketones (off-flavors). Hydrolytic rancidity: lipases/water hydrolyze ester bonds, releasing short-chain FAs like butyric acid (butter rancidity).

Q100. Waxes are esters of which two components?

A. Glycerol and fatty acids
B. Long-chain fatty acid and long-chain fatty alcohol
C. Sphingosine and fatty acid
D. Cholesterol and phosphate
E. Glycerol and phosphate

Answer: B Waxes are esters formed between a long-chain fatty acid (C14–C36) and a long-chain monohydric alcohol (C16–C30); they are solid at room temperature and serve protective/waterproofing roles.

SECTION U: LIPIDS-IV

Q101. Phosphatidylcholine (lecithin) is important in lung physiology because it is:

A. A precursor for steroid hormones
B. The major component of pulmonary surfactant (dipalmitoylphosphatidylcholine) that reduces alveolar surface tension
C. The primary lipid of myelin
D. An energy storage lipid
E. A structural component of nuclear membranes only

Answer: B Dipalmitoylphosphatidylcholine (DPPC) is the major component of surfactant produced by type II pneumocytes; its deficiency in premature neonates causes Respiratory Distress Syndrome (RDS).

Q102. Gangliosides differ from cerebrosides in that gangliosides:

A. Contain only one sugar residue
B. Contain one or more sialic acid (N-acetylneuraminic acid) residues attached to the oligosaccharide chain
C. Lack a ceramide backbone
D. Are found only in peripheral nerves
E. Contain phosphate groups

Answer: B Gangliosides are glycosphingolipids containing sialic acid (NANA); they are abundant in neuronal cell membranes and synapses. Cerebrosides have only neutral sugars (glucose or galactose), no sialic acid.

Q103. Tay-Sachs disease results from accumulation of which lipid?

A. Glucocerebroside (glucosylceramide)
B. Galactocebroside
C. GM2 ganglioside (due to hexosaminidase A deficiency)
D. Sphingomyelin
E. Ceramide

Answer: C Tay-Sachs is caused by deficiency of hexosaminidase A (lysosomal enzyme), leading to accumulation of GM2 ganglioside in neurons, causing progressive neurological deterioration and cherry-red spot.

Q104. In phospholipid structure, the sn-2 position of glycerophospholipids typically contains:

A. A saturated fatty acid
B. An unsaturated fatty acid
C. A phosphate group
D. A polar head group
E. A sphingosine backbone

Answer: B By convention, sn-1 position holds a saturated fatty acid and sn-2 holds an unsaturated fatty acid in most membrane glycerophospholipids — this asymmetry affects membrane fluidity and lipid signaling.

Q105. Phosphatidylinositol-4,5-bisphosphate (PIP₂) is important in cell signaling because:

A. It directly activates adenylyl cyclase
B. It is cleaved by phospholipase C to generate IP₃ (triggers Ca²⁺ release) and DAG (activates PKC)
C. It inhibits phospholipase A₂
D. It is the precursor for prostaglandins
E. It forms the major lipid of the inner mitochondrial membrane

Answer: B PLC cleaves PIP₂ into IP₃ (which releases Ca²⁺ from ER via IP₃ receptors) and DAG (which activates protein kinase C) — the phosphoinositide signaling pathway.

SECTION V: LIPIDS-V (EICOSANOIDS)

Q106. Aspirin irreversibly inhibits COX-1 and COX-2 by:

A. Competitive inhibition at the arachidonic acid binding site
B. Acetylation of a serine residue in the active site of cyclooxygenase
C. Blocking lipoxygenase
D. Inhibiting phospholipase A₂
E. Chelating the heme iron of COX

Answer: B Aspirin acetylates Ser530 of COX-1 and Ser516 of COX-2 — irreversible covalent modification that permanently inactivates the enzyme (unlike other NSAIDs, which are reversible inhibitors).

Q107. Leukotrienes (LTC₄, LTD₄, LTE₄) are derived from arachidonic acid via which enzyme and are primarily responsible for:

A. Thromboxane synthase; platelet aggregation
B. 5-Lipoxygenase; bronchoconstriction and increased vascular permeability in asthma/allergy
C. COX-1; gastric mucosal protection
D. Prostacyclin synthase; vasodilation
E. COX-2; fever and pain

Answer: B Cysteinyl leukotrienes (LTC₄, LTD₄, LTE₄) are potent bronchoconstrictors and mediators of allergic inflammation, 100–1000× more potent than histamine; synthesized via 5-LOX from arachidonic acid.

Q108. Prostacyclin (PGI₂) and thromboxane A₂ (TXA₂) have opposing effects. Which pairing is correct?

A. PGI₂: platelet aggregation; TXA₂: vasodilation
B. PGI₂: vasodilation + inhibits platelet aggregation; TXA₂: vasoconstriction + promotes platelet aggregation
C. Both cause vasoconstriction
D. Both inhibit platelet aggregation
E. PGI₂ is derived from COX-2 only; TXA₂ from COX-1 only

Answer: B PGI₂ (from endothelium) inhibits aggregation (↑cAMP in platelets) and causes vasodilation; TXA₂ (from platelets) promotes aggregation and vasoconstriction — a physiological balance regulating thrombosis.

Q109. Eicosanoids are derived from which 20-carbon polyunsaturated fatty acid?

A. Linoleic acid
B. Alpha-linolenic acid
C. Arachidonic acid
D. EPA (eicosapentaenoic acid) exclusively
E. Oleic acid

Answer: C Eicosanoids are primarily derived from arachidonic acid (20:4, ω-6) released from membrane phospholipids by phospholipase A₂; EPA can also serve as a precursor producing less inflammatory series-3 eicosanoids.

Q110. Zileuton, used in asthma management, exerts its effect by:

A. Blocking leukotriene receptors (CysLT1)
B. Inhibiting 5-lipoxygenase, reducing leukotriene synthesis
C. Inhibiting COX-2
D. Antagonizing thromboxane receptors
E. Inhibiting phospholipase A₂

Answer: B Zileuton directly inhibits 5-lipoxygenase, preventing conversion of arachidonic acid to leukotrienes. Montelukast and zafirlukast block CysLT1 receptors (different mechanism).

SECTION W: ENZYMES-I

Q111. According to the IUB classification system, an enzyme catalyzing the reaction: ATP + glucose → ADP + glucose-6-phosphate belongs to which class?

A. Oxidoreductase
B. Transferase
C. Hydrolase
D. Lyase
E. Isomerase

Answer: B Hexokinase transfers the gamma-phosphate group from ATP to glucose — transfer of a functional group from one molecule to another defines transferases (EC class 2).

Q112. Coenzyme NAD⁺ is derived from which vitamin?

A. Thiamine (B1)
B. Riboflavin (B2)
C. Niacin (B3)
D. Pyridoxine (B6)
E. Cobalamin (B12)

Answer: C NAD⁺ and NADP⁺ are derived from niacin (nicotinic acid/nicotinamide, Vitamin B3); deficiency causes pellagra (dermatitis, diarrhea, dementia).

Q113. An enzyme with a very low Km value indicates:

A. Low affinity for the substrate requiring high concentrations to work
B. High affinity for the substrate — half-maximal velocity is achieved at low substrate concentration
C. High Vmax
D. Susceptibility to competitive inhibition only
E. The enzyme is allosteric

Answer: B Km = substrate concentration at half-Vmax. Low Km means the enzyme achieves half-maximal rate at a low substrate concentration, indicating HIGH affinity for its substrate.

Q114. Pyridoxal phosphate (PLP) is the prosthetic group of which type of enzyme?

A. Oxidases
B. Aminotransferases (transaminases)
C. Carboxylases
D. Dehydrogenases
E. Kinases

Answer: B PLP (derived from vitamin B6) is the coenzyme for all aminotransferases (transaminases), forming a Schiff base with amino acids to transfer amino groups in transamination reactions.

Q115. Isoenzymes (isozymes) are:

A. Enzymes with identical structure from different species
B. Multiple molecular forms of the same enzyme activity, encoded by different genes or derived by different post-translational modifications, often with different kinetic/regulatory properties
C. Inhibited forms of the same enzyme
D. Enzymes with different activities encoded by the same gene
E. Enzymes that catalyze the reverse reaction

Answer: B Isozymes catalyze the same reaction but differ in structure and properties. Example: LDH has 5 isozymes (combinations of H and M subunits); LDH-1 predominates in heart and LDH-5 in liver — clinically useful in diagnosis.

SECTION X: ENZYMES-II

Q116. In competitive inhibition, increasing substrate concentration:

A. Cannot overcome inhibition regardless of substrate concentration
B. Can overcome inhibition — Vmax remains unchanged, but apparent Km increases
C. Decreases Vmax permanently
D. Has no effect on reaction velocity
E. Decreases both Km and Vmax

Answer: B Competitive inhibitors bind the same site as substrate; excess substrate can displace the inhibitor. Vmax is unchanged, but apparent Km increases (more substrate needed to achieve half-Vmax).

Q117. An enzyme displays sigmoidal kinetics rather than hyperbolic (Michaelis-Menten) kinetics. This indicates:

A. Non-enzymatic catalysis
B. Allosteric regulation with cooperative substrate binding (multiple active sites interact)
C. Irreversible inhibition
D. The enzyme is a monomeric hydrolase
E. The substrate concentration is below Km

Answer: B Sigmoidal kinetics (as in hemoglobin binding O₂ or ATCase) reflect cooperative allosteric behavior — binding at one subunit changes affinity at others, giving the characteristic S-shaped curve.

Q118. Statins inhibit HMG-CoA reductase by which mechanism?

A. Irreversible covalent modification of the active site
B. Competitive inhibition — structural analogs of HMG-CoA that bind the active site
C. Allosteric inhibition at a regulatory site distant from the active site
D. Non-competitive inhibition independent of substrate concentration
E. Uncompetitive inhibition by binding the enzyme-substrate complex

Answer: B Statins are structural analogs of the HMG-CoA transition state; they competitively and very tightly (but reversibly) bind the HMG-CoA reductase active site, blocking mevalonate synthesis.

Q119. The significance of Vmax in enzyme kinetics is that it represents:

A. The substrate concentration at half-maximal velocity
B. The maximum catalytic rate achieved when all enzyme active sites are saturated with substrate
C. The enzyme-substrate affinity constant
D. The rate of enzyme synthesis
E. The free energy of activation of the reaction

Answer: B Vmax is reached when all available enzyme molecules are in the enzyme-substrate complex (ES) — it reflects the total catalytic capacity and is directly proportional to total enzyme concentration.

Q120. Temperature increase beyond the optimal temperature causes enzyme activity to decrease because:

A. Substrate concentration decreases
B. The enzyme protein denatures — disruption of non-covalent bonds destroys the active site conformation
C. Km increases proportionally
D. Inhibitory products accumulate
E. Co-factor dissociation increases Vmax

Answer: B Beyond the optimum (~37°C in humans), thermal energy disrupts hydrogen bonds, hydrophobic interactions, and ionic bonds maintaining the active site geometry, causing irreversible denaturation.

SECTION Y: NUCLEIC ACID-I

Q121. The B-form of DNA (Watson-Crick model) is characterized by:

A. Left-handed helix with 11 bp per turn
B. Right-handed helix with ~10 bp per turn and major/minor grooves
C. Parallel strand orientation
D. RNA-like sugar pucker
E. Triple-stranded structure

Answer: B B-DNA is a right-handed double helix with ~10.5 bp/turn, base pairs inside (stacked perpendicular to the axis), and prominent major and minor grooves accessible for protein binding.

Q122. In DNA, adenine pairs with thymine via how many hydrogen bonds, while guanine pairs with cytosine via how many?

A. A-T: 3 H-bonds; G-C: 2 H-bonds
B. A-T: 2 H-bonds; G-C: 3 H-bonds
C. A-T: 1 H-bond; G-C: 2 H-bonds
D. A-T: 2 H-bonds; G-C: 2 H-bonds
E. A-T: 3 H-bonds; G-C: 3 H-bonds

Answer: B A-T forms 2 hydrogen bonds; G-C forms 3 hydrogen bonds — this is why G-C rich DNA has a higher melting temperature (Tm) than A-T rich DNA.

Q123. Which of the following is found in RNA but NOT in DNA?

A. Deoxyribose sugar and thymine
B. Ribose sugar and uracil
C. Adenine and guanine
D. Phosphodiester bonds
E. Purines and pyrimidines

Answer: B RNA contains ribose (2'-OH) instead of deoxyribose and uracil instead of thymine. DNA has deoxyribose and thymine.

Q124. Transfer RNA (tRNA) secondary structure includes:

A. A double helix identical to DNA
B. A cloverleaf structure with the anticodon loop, TΨC loop, DHU loop, and CCA-3' acceptor stem
C. A simple linear single strand
D. A triple helix with rRNA
E. A hairpin loop only

Answer: B tRNA folds into a characteristic cloverleaf secondary structure (and L-shaped tertiary structure) with 4 main loops: anticodon loop (decodes mRNA), DHU loop, TΨC loop, and the 3'-CCA acceptor end.

Q125. Chargaff's rules state that in double-stranded DNA:

A. A=G and T=C (purines equal pyrimidines in each type)
B. A=T and G=C (complementary base pairing — percentage of A equals T, and G equals C)
C. A=C and T=G
D. A+T = G+C in all organisms
E. A:T ratio = 2:1

Answer: B Chargaff's rules: [A]=[T] and [G]=[C] in any double-stranded DNA sample; also (A+G)=(T+C) — purines equal pyrimidines. This was key evidence leading to the Watson-Crick model.

SECTION Z: NUCLEIC ACID-II & III / GENETICS I–III

Q126. The enzyme that synthesizes RNA from a DNA template is:

A. DNA polymerase III
B. Primase
C. RNA polymerase (transcriptase)
D. Reverse transcriptase
E. Telomerase

Answer: C RNA polymerase reads the DNA template strand 3'→5' and synthesizes the RNA transcript 5'→3'. In prokaryotes, a single RNA pol handles all RNA types; eukaryotes have RNA pol I, II, and III.

Q127. During DNA replication, the lagging strand is synthesized:

A. Continuously in the 5'→3' direction toward the replication fork
B. Discontinuously as Okazaki fragments in the 5'→3' direction, away from the replication fork
C. Without the need for a primer
D. By DNA polymerase I in eukaryotes
E. In the 3'→5' direction

Answer: B The lagging strand template runs 5'→3' toward the fork, so polymerase (which can only add 5'→3') must synthesize it in short Okazaki fragments that are later joined by DNA ligase.

Q128. Rifampicin inhibits transcription in bacteria by:

A. Intercalating into DNA
B. Binding the beta subunit of bacterial RNA polymerase, blocking the initiation of RNA synthesis
C. Inhibiting the ribosome 30S subunit
D. Inhibiting DNA gyrase
E. Degrading mRNA after transcription

Answer: B Rifampicin binds the β-subunit of bacterial RNA polymerase, blocking the initiation step of transcription. It is used against Mycobacterium tuberculosis and Mycobacterium leprae.

Q129. The genetic code is described as "degenerate." This means:

A. Some codons code for more than one amino acid
B. Most amino acids are encoded by more than one codon (synonymous codons)
C. Some amino acids have no codon
D. The code varies between species
E. Codons overlap with each other

Answer: B Degeneracy means that most of the 20 amino acids are specified by more than one codon (64 codons encode only 20 AAs + 3 stop codons). The wobble typically occurs at the 3rd codon position.

Q130. Which of the following is NOT required for translation (protein synthesis)?

A. mRNA template
B. Ribosomes (40S + 60S in eukaryotes)
C. tRNA aminoacylated by aminoacyl-tRNA synthetase
D. GTP (for elongation factor EF-Tu/eEF1A and translocation)
E. DNA polymerase

Answer: E Translation requires mRNA, ribosomes, charged tRNAs, initiation/elongation/termination factors, GTP, and ATP. DNA polymerase is used in DNA replication, not in translation.

ANSWER KEY SUMMARY

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	C	27	C	53	B	79	A	105	B
2	B	28	C	54	B	80	B	106	B
3	C	29	D	55	B	81	B	107	B
4	C	30	B	56	B	82	C	108	B
5	B	31	A	57	C	83	B	109	C
6	B	32	C	58	B	84	A	110	B
7	B	33	C	59	B	85	C	111	B
8	B	34	B	60	B	86	C	112	C
9	B	35	B	61	C	87	C	113	B
10	B	36	C	62	B	88	B	114	B
11	B	37	C	63	B	89	C	115	B
12	C	38	C	64	C	90	B	116	B
13	B	39	B	65	B	91	B	117	B
14	B	40	C	66	B	92	B	118	B
15	B	41	B	67	B	93	B	119	B
16	D	42	B	68	B	94	B	120	B
17	D	43	B	69	B	95	C	121	B
18	C	44	C	70	D	96	C	122	B
19	B	45	C	71	D	97	B	123	B
20	C	46	B	72	C	98	B	124	B
21	C	47	B	73	B	99	B	125	B
22	C	48	C	74	B	100	B	126	C
23	B	49	B	75	C	101	B	127	B
24	B	50	B	76	B	102	B	128	B
25	B	51	C	77	C	103	C	129	B
26	C	52	B	78	C	104	B	130	E

130 MCQs | 26 Topics | All answers with explanations

Topics covered: Cell → Cell Membrane → Membrane Transport → pH & Buffers → Carbohydrates (Mono/Di/Polysaccharides) → Proteins (I–IV, Structure I–II) → Lipids (I–V) → Enzymes (I–II) → Nucleic Acids (I–III) → Genetics (I–III)

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Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	C	27	C	53	B	79	A	105	B
2	B	28	C	54	B	80	B	106	B
3	C	29	D	55	B	81	B	107	B
4	C	30	B	56	B	82	C	108	B
5	B	31	A	57	C	83	B	109	C
6	B	32	C	58	B	84	A	110	B
7	B	33	C	59	B	85	C	111	B
8	B	34	B	60	B	86	C	112	C
9	B	35	B	61	C	87	C	113	B
10	B	36	C	62	B	88	B	114	B
11	B	37	C	63	B	89	C	115	B
12	C	38	C	64	C	90	B	116	B
13	B	39	B	65	B	91	B	117	B
14	B	40	C	66	B	92	B	118	B
15	B	41	B	67	B	93	B	119	B
16	D	42	B	68	B	94	B	120	B
17	D	43	B	69	B	95	C	121	B
18	C	44	C	70	D	96	C	122	B
19	B	45	C	71	D	97	B	123	B
20	C	46	B	72	C	98	B	124	B
21	C	47	B	73	B	99	B	125	B
22	C	48	C	74	B	100	B	126	C
23	B	49	B	75	C	101	B	127	B
24	B	50	B	76	B	102	B	128	B
25	B	51	C	77	C	103	C	129	B
26	C	52	B	78	C	104	B	130	E

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	C	27	C	53	B	79	A	105	B
2	B	28	C	54	B	80	B	106	B
3	C	29	D	55	B	81	B	107	B
4	C	30	B	56	B	82	C	108	B
5	B	31	A	57	C	83	B	109	C
6	B	32	C	58	B	84	A	110	B
7	B	33	C	59	B	85	C	111	B
8	B	34	B	60	B	86	C	112	C
9	B	35	B	61	C	87	C	113	B
10	B	36	C	62	B	88	B	114	B
11	B	37	C	63	B	89	C	115	B
12	C	38	C	64	C	90	B	116	B
13	B	39	B	65	B	91	B	117	B
14	B	40	C	66	B	92	B	118	B
15	B	41	B	67	B	93	B	119	B
16	D	42	B	68	B	94	B	120	B
17	D	43	B	69	B	95	C	121	B
18	C	44	C	70	D	96	C	122	B
19	B	45	C	71	D	97	B	123	B
20	C	46	B	72	C	98	B	124	B
21	C	47	B	73	B	99	B	125	B
22	C	48	C	74	B	100	B	126	C
23	B	49	B	75	C	101	B	127	B
24	B	50	B	76	B	102	B	128	B
25	B	51	C	77	C	103	C	129	B
26	C	52	B	78	C	104	B	130	E

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	C	27	C	53	B	79	A	105	B
2	B	28	C	54	B	80	B	106	B
3	C	29	D	55	B	81	B	107	B
4	C	30	B	56	B	82	C	108	B
5	B	31	A	57	C	83	B	109	C
6	B	32	C	58	B	84	A	110	B
7	B	33	C	59	B	85	C	111	B
8	B	34	B	60	B	86	C	112	C
9	B	35	B	61	C	87	C	113	B
10	B	36	C	62	B	88	B	114	B
11	B	37	C	63	B	89	C	115	B
12	C	38	C	64	C	90	B	116	B
13	B	39	B	65	B	91	B	117	B
14	B	40	C	66	B	92	B	118	B
15	B	41	B	67	B	93	B	119	B
16	D	42	B	68	B	94	B	120	B
17	D	43	B	69	B	95	C	121	B
18	C	44	C	70	D	96	C	122	B
19	B	45	C	71	D	97	B	123	B
20	C	46	B	72	C	98	B	124	B
21	C	47	B	73	B	99	B	125	B
22	C	48	C	74	B	100	B	126	C
23	B	49	B	75	C	101	B	127	B
24	B	50	B	76	B	102	B	128	B
25	B	51	C	77	C	103	C	129	B
26	C	52	B	78	C	104	B	130	E