Chapter 98 – MCQs for Lab (Biochemistry)

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Chapter 98 of 5000 Plus MCQs for Lab Technician and Technologists are designed to test the knowledge and proficiency of laboratory professionals who work in the field of clinical laboratory science.

4851 to 4900 MCQs for Lab Technician and Technologist Exam Preparation

5000 Plus MCQs for Lab Technician and Technologists are designed to test the knowledge and proficiency of laboratory professionals who work in the field of clinical laboratory science. These questions cover a wide range of topics related to laboratory science, including anatomy, physiology, microbiology, chemistry, and hematology.

MCQs Chapter 98 (Lab)
MCQs Chapter 98 (Lab)

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Questions 4851 to 4900

Which of the following reactions is unique to gluconeogenesis?
  • a) Lactate → Pyruvate
  • b) Phosphoenolpyruvate → Pyruvate
  • c) Oxaloacetate → Phosphoenolpyruvate
  • d) Glucose-6-phosphate → Fructose-6-phosphate
Explanation: The conversion of Oxaloacetate to Phosphoenolpyruvate (catalyzed by PEPCK) is one of the key bypass reactions unique to gluconeogenesis. It allows the cell to bypass the irreversible step of glycolysis catalyzed by pyruvate kinase.
Why other options are incorrect:
  • • Lactate → Pyruvate: This is a reversible reaction catalyzed by lactate dehydrogenase and is not exclusive to gluconeogenesis.
  • • PEP → Pyruvate: This reaction is actually the final step of glycolysis (catalyzed by pyruvate kinase) and is irreversible in that direction.
  • • G6P → F6P: This is a reversible isomerization reaction shared by both glycolysis and gluconeogenesis pathways.
The synthesis of glucose from pyruvate by gluconeogenesis:
  • a) Requires the participation of biotin
  • b) Occurs exclusively in the cytosol
  • c) Is inhibited by elevated level of insulin
  • d) Requires oxidation/reduction of FAD
Explanation: The first step of gluconeogenesis is the conversion of pyruvate to oxaloacetate by the enzyme pyruvate carboxylase. This enzyme requires biotin as a CO2 carrier to function.
Why other options are incorrect:
  • • Occurs exclusively in the cytosol: Incorrect. It begins in the mitochondria (pyruvate to oxaloacetate) and involves the endoplasmic reticulum (glucose-6-phosphatase step).
  • • Is inhibited by elevated level of insulin: While physiologically true (insulin suppresses gluconeogenic enzymes), Option A is the fundamental biochemical requirement for the reaction pathway to occur.
  • • Requires oxidation/reduction of FAD: Incorrect. Gluconeogenesis requires NADH (for the reduction of 1,3-bisphosphoglycerate), not FAD.
The conversion of pyruvate to acetyl CoA and CO2:
  • a) Is reversible
  • b) Involves the participation of lipoic acid
  • c) Depends on the coenzyme biotin
  • d) Occurs in the cytosol
Explanation: The conversion of pyruvate to acetyl CoA is catalyzed by the Pyruvate Dehydrogenase Complex (PDC). This multi-enzyme complex requires five cofactors: TPP, Lipoic acid, CoA, FAD, and NAD+.
Why other options are incorrect:
  • • Is reversible: This reaction is a highly exergonic, irreversible oxidative decarboxylation in animal cells.
  • • Depends on the coenzyme biotin: Biotin is required for carboxylation reactions (like pyruvate to oxaloacetate), not decarboxylation.
  • • Occurs in the cytosol: Pyruvate must be transported into the mitochondrial matrix for this reaction to take place.
Pasteur effect is
  • a) Inhibition of glycolysis
  • b) Oxygen is involved
  • c) Inhibition of enzyme phosphofructokinase
  • d) All of these
Explanation: The Pasteur effect refers to the inhibition of glycolysis in the presence of oxygen. Oxygen promotes oxidative phosphorylation, increasing ATP production, which inhibits phosphofructokinase-1 (PFK-1) and reduces the rate of glycolysis. Therefore, all of the given statements are correct.
Why other options are incorrect:
  • • Inhibition of glycolysis: This is the main feature of the Pasteur effect.
  • • Oxygen is involved: The effect occurs when oxygen becomes available and aerobic metabolism is favored.
  • • Inhibition of enzyme phosphofructokinase: Increased ATP availability inhibits PFK-1, contributing to reduced glycolysis.
How many ATPs are produced in the conversion of phosphoenol pyruvate to citrate?
  • a) 1
  • b) 2
  • c) 4
  • d) 6
Explanation: The conversion of Phosphoenolpyruvate (PEP) to Citrate involves two main ATP-generating steps:
  1. PEP to Pyruvate: Produces 1 ATP directly via substrate-level phosphorylation (catalyzed by Pyruvate Kinase).
  2. Pyruvate to Acetyl-CoA: Produces 1 NADH via the Pyruvate Dehydrogenase complex. In the electron transport chain, 1 NADH yields approximately 3 ATP (classic calculation).
  3. Acetyl-CoA to Citrate: This condensation reaction does not produce ATP.
Total: 1 (direct) + 3 (from NADH) = 4 ATP.
Why other options are incorrect:
  • • 1: This only accounts for the substrate-level phosphorylation and ignores the oxidative phosphorylation from NADH.
  • • 2: There is no standard metabolic calculation for this specific pathway that results in only 2 ATP.
  • • 6: This would be the yield for a different sequence or for two moles of PEP; for a single molecule, the yield is 4.
Reduced glutathione functions in R.B.Cs to:
  • a) Produce NADPH
  • b) Reduce methemoglobin to hemoglobin
  • c) Produce NADH
  • d) Reduce oxidizing agents such as H2O2
Explanation: The primary role of Reduced Glutathione (GSH) in Red Blood Cells is to protect the cell from oxidative damage. It neutralizes reactive oxygen species like hydrogen peroxide (H2O2) via the enzyme glutathione peroxidase, preventing the oxidation of hemoglobin and membrane lipids.
Why other options are incorrect:
  • • Produce NADPH: Incorrect. It is the HMP Shunt (Pentose Phosphate Pathway) that produces NADPH, which is then used to regenerate reduced glutathione.
  • • Reduce methemoglobin to hemoglobin: While glutathione can help slightly, this is primarily the job of the enzyme NADH-methemoglobin reductase (Diaphorase).
  • • Produce NADH: Incorrect. NADH is produced during Glycolysis (Embden-Meyerhof pathway).
Phenylalanine is the precursor of:
  • a) L-DOPA
  • b) Histamine
  • c) Tyrosine
  • d) Thyroxine
Explanation: Phenylalanine is an essential amino acid that is converted directly into Tyrosine by the enzyme phenylalanine hydroxylase. This is the primary metabolic route for phenylalanine.
Why other options are incorrect:
  • • L-DOPA: While phenylalanine eventually becomes L-DOPA, Tyrosine is the immediate precursor of L-DOPA.
  • • Histamine: Histamine is synthesized via the decarboxylation of the amino acid Histidine, not phenylalanine.
  • • Thyroxine: Similar to L-DOPA, Thyroxine is synthesized from Tyrosine (after it is incorporated into thyroglobulin).
D-Mannose is present in some plant products like:
  • a) Resins
  • b) Pectins
  • c) Mucilage
  • d) Gums
Explanation: D-Mannose is an aldohexose sugar that is widely distributed in plants, usually as a component of complex polysaccharides. It is a major constituent of many plant Gums (such as Guar gum and Locust bean gum) and hemicelluloses (mannans).
Why other options are incorrect:
  • • Resins: These are sticky organic substances (like pine resin) composed mainly of terpenes and resin acids, not sugars.
  • • Pectins: These are structural heteropolysaccharides found in primary cell walls, composed primarily of galacturonic acid.
  • • Mucilage: While some mucilages contain mannose, they are more typically characterized by sugars like galactose, arabinose, and xylose. Gums are the more prominent source for mannose-based polymers.
Galactose is a main constituent of:
  • a) Milk sugar
  • b) Honey
  • c) Cane sugar
  • d) Chitin
Explanation: Galactose is a monosaccharide that, when combined with glucose via a beta-1,4-glycosidic bond, forms the disaccharide Lactose, which is the primary sugar found in milk (**Milk sugar**).
Why other options are incorrect:
  • • Honey: Primarily consists of a mixture of the monosaccharides Fructose and Glucose.
  • • Cane sugar: Also known as Sucrose, it is a disaccharide composed of Glucose and Fructose.
  • • Chitin: A structural polysaccharide found in fungal cell walls and arthropod exoskeletons, composed of N-acetylglucosamine units.
Glucosamine is an important constituent of:
  • a) Homopolysaccharide
  • b) Heteropolysaccharide
  • c) Mucopolysaccharide
  • d) Dextran
Explanation: Glucosamine is an amino sugar that serves as a primary building block for Mucopolysaccharides (also known as Glycosaminoglycans or GAGs), such as hyaluronic acid, heparin, and keratan sulfate.
Why other options are incorrect:
  • • Homopolysaccharide: These consist of only one type of monosaccharide (e.g., Starch or Cellulose). Glucosamine-containing polymers usually involve alternating units.
  • • Heteropolysaccharide: While Mucopolysaccharides are technically a type of heteropolysaccharide, “Mucopolysaccharide” is the more specific and clinically relevant classification for glucosamine’s role in connective tissues.
  • • Dextran: This is a complex branched glucan (homopolysaccharide) composed entirely of Glucose units.
Glycogen is present in all body tissues except:
  • a) Liver
  • b) Brain
  • c) Kidney
  • d) Stomach
Explanation: While glycogen is the primary storage form of glucose in the body, its distribution is not equal. The Brain has virtually no significant glycogen stores (it contains only negligible amounts in astrocytes). Because of this, the brain is highly sensitive to hypoglycemia and requires a continuous supply of glucose from the blood.
Why other options are incorrect:
  • • Liver: This is the major site of glycogen storage (up to 10% of liver weight) used to maintain blood glucose levels.
  • • Kidney: Small amounts of glycogen are stored in the renal cortex and medulla to provide energy for renal functions.
  • • Stomach: The smooth muscles and mucosal lining of the stomach contain glycogen stores to power muscular contractions and cellular activity.
Iodine test is positive for starch, dextrin and:
  • a) Mucoproteins
  • b) Agar
  • c) Glycogen
  • d) Cellulose
Explanation: The Iodine test is used to detect polysaccharides. It works because the iodine molecules get trapped inside the helical structure of alpha-glucans. Glycogen reacts with iodine to give a characteristic reddish-brown (mahogany) color.
Why other options are incorrect:
  • • Mucoproteins: These are proteins conjugated with carbohydrates; they do not possess the specific helical glucose chains required to trap iodine.
  • • Agar: Agar is a polymer of galactose (sulfated galactans), which does not form the necessary structure to give a positive iodine test.
  • • Cellulose: Although it is a glucose polymer, it has beta-1,4-linkages that form straight, rigid fibers rather than the helices found in starch or glycogen, so it does not trap iodine.
The general formula for polysaccharide is:
  • a) (C6H10O5)n
  • b) (C6H12O6)n
  • c) (C6H12O5)n
  • d) (C5H10O5)n
Explanation: Polysaccharides are formed by the condensation of many monosaccharide units (usually hexoses like glucose, C6H12O6). During the formation of each glycosidic bond, one molecule of water (H2O) is removed. Therefore, the formula for each unit becomes C6H10O5 (C6H12O6 minus H2O).
Why other options are incorrect:
  • • (C6H12O6)n: This is the formula for a hexose monosaccharide. It doesn’t account for the loss of water during polymerization.
  • • (C6H12O5)n: This formula has too many hydrogen atoms and does not represent a standard carbohydrate polymer.
  • • (C5H10O5)n: This represents a polymer of pentose sugars (like ribose) where no water has been removed, which is also incorrect for a finished polysaccharide chain.
Epimers of glucose is:
  • a) Fructose
  • b) Galactose
  • c) Ribose
  • d) Deoxyribose
Explanation: Epimers are stereoisomers that differ in configuration at only one specific carbon atom. Galactose is a C-4 epimer of glucose, meaning they are identical except for the orientation of the hydroxyl (-OH) group at the fourth carbon.
Why other options are incorrect:
  • • Fructose: This is a functional isomer of glucose (keto-aldo isomerism), not an epimer, as it has a ketone group instead of an aldehyde group.
  • • Ribose: This is a pentose sugar (5 carbons), whereas glucose is a hexose (6 carbons). Epimers must have the same number of carbon atoms.
  • • Deoxyribose: This is also a pentose sugar and is missing an oxygen atom at the C-2 position, making it a “deoxy” sugar rather than an epimer of glucose.
Human heart muscle contains:
  • a) D-Arabinose
  • b) D-Ribose
  • c) D-Xylose
  • d) L-Xylose
Explanation: D-Ribose is a pentose sugar that is a fundamental component of nucleotides like ATP (Adenosine Triphosphate), as well as RNA. Since heart muscle is highly metabolic and requires a constant, high supply of energy, it contains significant amounts of D-Ribose within its energy-carrying molecules.
Why other options are incorrect:
  • • D-Arabinose: Chiefly found in plant gums and bacterial cell walls; it is not a structural component of human muscle.
  • • D-Xylose: Also known as “wood sugar,” it is found in plant hemicelluloses and is not stored in human heart tissue.
  • • L-Xylose: This is a rare sugar that is excreted in the urine of individuals with the metabolic condition essential pentosuria, but it is not a normal constituent of heart muscle.
The intermediate in hexose monophosphate shunt is:
  • a) D-Ribulose
  • b) D-Arabinose
  • c) D-Xylose
  • d) D-Lyxose
Explanation: The Hexose Monophosphate (HMP) Shunt (or Pentose Phosphate Pathway) produces several pentose sugars as intermediates. D-Ribulose (specifically as D-Ribulose-5-phosphate) is produced during the oxidative phase of the pathway by the decarboxylation of 6-phosphogluconate.
Why other options are incorrect:
  • • D-Arabinose: This is a pentose sugar found mainly in plants and bacteria, not a standard intermediate of human glucose metabolism.
  • • D-Xylose: While Xylulose-5-phosphate is an intermediate in the HMP shunt, Xylose itself is not.
  • • D-Lyxose: This is an aldopentose sugar that is very rare in nature and does not play a role in the human HMP shunt pathway.
On boiling, Benedict’s solution is NOT reduced by:
  • a) Sucrose
  • b) Lactose
  • c) Maltose
  • d) Fructose
Explanation: Benedict’s solution is used to detect reducing sugars. Sucrose is a non-reducing sugar because its constituent monosaccharides (glucose and fructose) are linked by their anomeric carbons, meaning there is no free aldehyde or ketone group available to reduce the copper in the reagent.
Why other options are incorrect:
  • • Lactose: A reducing disaccharide because it has a free hemiacetal group on the glucose unit.
  • • Maltose: A reducing disaccharide containing two glucose units, one of which has a free anomeric carbon.
  • • Fructose: Although a ketose, fructose is a reducing monosaccharide because it tautomerizes to an aldose under the alkaline conditions of Benedict’s reagent.
The distinguishing test between monosaccharides and disaccharide is:
  • a) Bial’s test
  • b) Seliwanoff’s test
  • c) Barfoed’s test
  • d) Hydrolysis test
Explanation: Barfoed’s test is used to distinguish monosaccharides from reducing disaccharides. It uses copper acetate in an acidic medium. Monosaccharides are stronger reducing agents and will react to form a red precipitate within 1–2 minutes, whereas reducing disaccharides react much more slowly (usually taking 7–12 minutes).
Why other options are incorrect:
  • • Bial’s test: Used specifically to distinguish Pentoses from Hexoses.
  • • Seliwanoff’s test: Used to distinguish Ketoses (like fructose) from Aldoses.
  • • Hydrolysis test: This is a general procedure to break down complex carbohydrates into simpler ones, not a specific diagnostic test to distinguish between the two classes.
Barfoed’s solution is NOT reduced by:
  • a) Glucose
  • b) Mannose
  • c) Sucrose
  • d) Ribose
Explanation: Sucrose is a non-reducing sugar. In Barfoed’s test, the reagent (copper acetate in acetic acid) requires a free aldehyde or ketone group to reduce the cupric ions to cuprous oxide. Since the anomeric carbons of both glucose and fructose are locked in the glycosidic bond in sucrose, it cannot reduce the solution.
Why other options are incorrect:
  • • Glucose: A monosaccharide that reduces Barfoed’s solution quickly (within 1-2 minutes).
  • • Mannose: An aldohexose monosaccharide that gives a positive result in Barfoed’s test.
  • • Ribose: An aldopentose monosaccharide that also reduces Barfoed’s solution rapidly.
Cori cycle is:
  • a) Synthesis of glucose
  • b) reuse of glucose
  • c) uptake of glucose
  • d) Both (Synthesis of glucose) & (reuse of glucose)
Explanation: The Cori Cycle (also known as the Lactic Acid Cycle) is a metabolic pathway in which lactate produced by anaerobic glycolysis in the muscles moves to the liver and is converted back into glucose (synthesis). This glucose then returns to the muscles to be used again for energy (reuse).
Why other options are incorrect:
  • • Synthesis of glucose: While true (gluconeogenesis occurs in the liver during this cycle), it only describes half of the process.
  • • Reuse of glucose: While the goal is to recycle lactate back into usable glucose, this phrase alone doesn’t capture the chemical synthesis part of the cycle.
  • • Uptake of glucose: This refers to the movement of glucose into cells (via GLUT transporters) but is a general physiological process, not a definition of the Cori cycle itself.
Cane sugar is known as:
  • a) Galactose
  • b) Sucrose
  • c) Fructose
  • d) Maltose
Explanation: Sucrose is commonly known as cane sugar or table sugar. It is a disaccharide composed of one molecule of glucose and one molecule of fructose, and it is primarily extracted from sugar cane and sugar beets.
Why other options are incorrect:
  • • Galactose: Known as “brain sugar,” it is a monosaccharide usually found as a component of lactose (milk sugar).
  • • Fructose: Known as “fruit sugar,” it is the sweetest naturally occurring monosaccharide found in fruits and honey.
  • • Maltose: Known as “malt sugar,” it is a disaccharide produced during the breakdown of starch.
Which of the following is NOT a reducing sugar?
  • a) Lactose
  • b) Maltose
  • c) Sucrose
  • d) Fructose
Explanation: Sucrose is a non-reducing sugar because the glycosidic bond is formed between the anomeric carbon of glucose (C1) and the anomeric carbon of fructose (C2). This means there is no free aldehyde or ketone group available to reduce metallic ions like Cu²⁺.
Why other options are incorrect:
  • • Lactose: A reducing disaccharide; it has one free anomeric carbon on the glucose unit.
  • • Maltose: A reducing disaccharide; it has one free anomeric carbon on the second glucose unit.
  • • Fructose: All monosaccharides are reducing sugars. Even though fructose is a ketose, it can tautomerize to an aldose in alkaline solutions to act as a reducing agent.
α-D-Glucose and β-D-glucose are related by:
  • a) Epimers
  • b) Anomers
  • c) Multirotation
  • d) Ketoenol pair
Explanation: Anomers are a special type of stereoisomer that differ in configuration only at the hemiacetal or acetal carbon (C-1 for glucose), known as the anomeric carbon. In the α form, the hydroxyl group at C-1 is on the opposite side of the CH2OH group, while in the β form, they are on the same side.
Why other options are incorrect:
  • • Epimers: While anomers are technically a sub-type of epimers, “Anomers” is the more specific and scientifically accurate term for differences at the C-1 position of a cyclic sugar.
  • • Multirotation: (Usually spelled Mutarotation) This is the process of the change in optical rotation as α and β forms reach equilibrium in solution; it is not the name of the relationship between the molecules.
  • • Ketoenol pair: This refers to the interconversion between a ketone/aldehyde and an enol (e.g., glucose and fructose in alkaline solution), which involves a double bond.
The stable ring formation in D-Glucose involves:
  • a) C-1 and C-4
  • b) C-1 and C-2
  • c) C-1 and C-5
  • d) C-2 and C-5
Explanation: The most stable form of D-Glucose in solution is the pyranose ring. This six-membered ring is formed by an intramolecular reaction between the aldehyde group at C-1 and the hydroxyl (-OH) group at C-5.
Why other options are incorrect:
  • • C-1 and C-4: This would form a five-membered (furanose) ring. While possible, it is much less stable and less common for glucose than the pyranose form.
  • • C-1 and C-2: This does not form a stable cyclic hemiacetal ring due to the high strain of a three-membered ring.
  • • C-2 and C-5: This combination is not involved in the standard cyclization of D-glucose, as the hemiacetal bond must involve the carbonyl carbon (C-1).
Reduction of Glucose with Ca++ in water produces:
  • a) Sorbitol
  • b) Dulcitol
  • c) Mannitol
  • d) Glucuronic acid
Explanation: The reduction of the aldehyde group (-CHO) at the C-1 position of D-Glucose yields a polyhydric alcohol (sugar alcohol) known as Sorbitol (also called Glucitol).
Why other options are incorrect:
  • • Dulcitol: This is the sugar alcohol produced by the reduction of Galactose.
  • • Mannitol: This is produced by the reduction of Mannose or as a co-product of Fructose reduction.
  • • Glucuronic acid: This is an oxidation product of glucose (where the C-6 hydroxyl group is oxidized), not a reduction product.
Starch and glycogen are polymers of:
  • a) Fructose
  • b) Mannose
  • c) α−D-Glucose
  • d) Galactose
Explanation: Both starch (the storage form of energy in plants) and glycogen (the storage form of energy in animals) are polysaccharides composed entirely of α−D-Glucose units linked by glycosidic bonds.
Why other options are incorrect:
  • • Fructose: Forms the polymer Inulin, but is not a component of starch or glycogen.
  • • Mannose: An epimer of glucose; it forms mannans but not storage polymers like starch.
  • • Galactose: A component of lactose (milk sugar) and agar, but not the building block for starch or glycogen.
Reducing ability of carbohydrates is due to:
  • a) Carboxyl group
  • b) Hydroxyl group
  • c) Enediol formation
  • d) Ring structure
Explanation: In alkaline solutions (like Benedict’s or Fehling’s reagents), reducing sugars undergo tautomerization to form enediols. These enediol intermediates are highly reactive and act as powerful reducing agents, capable of reducing metal ions like $Cu^{2+}$ to $Cu^+$.
Why other options are incorrect:
  • • Carboxyl group: This is an acidic group (COOH). Carbohydrates contain carbonyl (aldehyde/ketone) and hydroxyl groups, not carboxyl groups (unless oxidized).
  • • Hydroxyl group: While all carbohydrates have -OH groups, the hydroxyl group alone does not provide the specific reducing power needed to reduce metallic salts.
  • • Ring structure: The stable cyclic form actually masks the reducing group; the ring must open to form the linear intermediate or enediol to exhibit reducing properties.
Which of the following is not a polymer of glucose?
  • a) Amylose
  • b) Inulin
  • c) Cellulose
  • d) Dextrin
Explanation: Inulin is a polymer of fructose (a fructan), not glucose. It is commonly found in the roots or rhizomes of plants like chicory and dahlia, and it is used clinically to measure the Glomerular Filtration Rate (GFR).
Why other options are incorrect:
  • • Amylose: A linear polymer composed entirely of glucose units linked by α(1→4) glycosidic bonds; it is a component of starch.
  • • Cellulose: A structural polysaccharide made of a long chain of glucose units linked by β(1→4) glycosidic bonds.
  • • Dextrin: A group of low-molecular-weight carbohydrates produced by the hydrolysis of starch, consisting entirely of glucose units.
Invert sugar is:
  • a) Lactose
  • b) Mannose
  • c) Fructose
  • d) Hydrolytic product of sucrose
Explanation: Invert sugar is an equimolar mixture of glucose and fructose obtained by the hydrolysis of sucrose. It is called “invert” because the optical rotation of the solution “inverts” from dextrorotatory (sucrose) to levorotatory (the mixture) during the process.
Why other options are incorrect:
  • • Lactose: A disaccharide found in milk composed of glucose and galactose; its hydrolysis does not produce invert sugar.
  • • Mannose: An aldose sugar and a C-2 epimer of glucose; it is a single monosaccharide, not a mixture.
  • • Fructose: While fructose is a component of invert sugar, it alone is not called invert sugar. Invert sugar specifically refers to the resulting mixture of glucose and fructose.
The carbohydrate reserved in human body is:
  • a) Starch
  • b) Glucose
  • c) Glycogen
  • d) Inulin
Explanation: Glycogen is the primary storage polysaccharide in humans and other vertebrates. It is highly branched and stored mainly in the liver and skeletal muscle tissues to be converted back into glucose when energy is needed.
Why other options are incorrect:
  • • Starch: This is the energy storage carbohydrate for plants. Humans consume it, but do not store it in this form.
  • • Glucose: This is the immediate “fuel” found in the bloodstream, but it is not a “reserve” because high levels are osmotic and toxic; hence, the body converts it to glycogen for storage.
  • • Inulin: A plant-derived polysaccharide (fructan) used by plants like dahlias for storage; the human body cannot produce it.
A disaccharide linked by α-1-4 Glycosidic linkages is:
  • a) Lactose
  • b) Sucrose
  • c) Cellulose
  • d) Maltose
Explanation: Maltose (malt sugar) is a disaccharide formed from two units of glucose joined with an α(1→4) glycosidic bond. It is produced during the digestion of starch by amylase.
Why other options are incorrect:
  • • Lactose: Composed of galactose and glucose linked by a β(1→4) glycosidic bond.
  • • Sucrose: Composed of glucose and fructose linked by an α1-β2 glycosidic bond.
  • • Cellulose: This is a polysaccharide (not a disaccharide) consisting of glucose units linked by β(1→4) glycosidic bonds.
All proteins contain the:
  • a) Same 20 amino acids
  • b) Different amino acids
  • c) 300 Amino acids occurring in nature
  • d) Only a few amino acids
Explanation: Despite the vast diversity of life, almost all proteins in all living organisms are constructed from the same 20 standard amino acids. These are the “building blocks” encoded by the universal genetic code.
Why other options are incorrect:
  • • Different amino acids: While the sequence and combination of amino acids differ between proteins, the set of available building blocks remains the same 20.
  • • 300 Amino acids occurring in nature: There are indeed over 300 amino acids found in nature, but the majority of these are “non-protein” amino acids and are not used as building blocks for protein synthesis.
  • • Only a few amino acids: “Few” is an underestimate; the complex structure of proteins requires the specific chemical diversity provided by the standard 20 amino acids.
Proteins contain:
  • a) Only L- α – amino acids
  • b) Only D-amino acids
  • c) DL-Amino acids
  • d) Both (Only L- α – amino acids) and (Only D-amino acids)
Explanation: Almost all proteins found in nature are composed exclusively of L-α-amino acids. While D-amino acids do exist in nature (e.g., in bacterial cell walls and some antibiotics), they are not used by ribosomes to build proteins.
Why other options are incorrect:
  • • Only D-amino acids: These are not found in the primary structure of proteins synthesized by living cells.
  • • DL-Amino acids: This refers to a racemic mixture (50/50) of both isomers. Biological systems are stereospecific and choose only the L-form for protein synthesis.
  • • Option (d): This is incorrect because proteins are not a mix of both types; they are strictly made of L-isomers to ensure proper folding and enzyme-substrate fitting.
The optically inactive amino acid is:
  • a) Glycine
  • b) Serine
  • c) Threonine
  • d) Valine
Explanation: Optical activity requires a chiral center (a carbon atom bonded to four different groups). **Glycine** is the simplest amino acid and its side chain (R-group) is a Hydrogen atom. Since the alpha-carbon is bonded to two identical hydrogen atoms, it is achiral and therefore optically inactive.
Why other options are incorrect:
  • • Serine: Has a -CH₂OH side chain, making the alpha-carbon chiral and the molecule optically active.
  • • Threonine: Contains two chiral centers, making it optically active.
  • • Valine: Has an isopropyl side chain, making the alpha-carbon chiral and the molecule optically active.
At neutral pH, a mixture of amino acids in solution would be predominantly:
  • a) Dipolar ions
  • b) Nonpolar molecules
  • c) Positive and monovalent
  • d) Hydrophobic
  • e) Negatively charged side chains
Explanation: At neutral pH, amino acids exist as Dipolar ions (also known as Zwitterions). In this state, the amino group is protonated ($-NH_3^+$) and the carboxyl group is deprotonated ($-COO^-$), giving the molecule both a positive and a negative charge while remaining electrically neutral overall.
Why other options are incorrect:
  • • Nonpolar molecules: Amino acids are inherently polar because of their charged amino and carboxyl groups.
  • • Positive and monovalent: This state occurs only in highly acidic (low pH) conditions where the carboxyl group is protonated ($-COOH$).
  • • Hydrophobic: This refers to the property of specific side chains (like Leucine or Valine), but does not describe the ionic state of the amino acid backbone in solution.
  • • Negatively charged side chains: Only acidic amino acids (like Aspartate and Glutamate) have negatively charged side chains at neutral pH; this is not a property of all amino acids.
The pH (isoelectric pH) of alanine is:
  • a) 6.02
  • b) 6.6
  • c) 6.8
  • d) 7.2
Explanation: The isoelectric point (pI) is the pH at which an amino acid has a net electrical charge of zero (zwitterion form). For Alanine, which has a carboxyl $pK_1$ of ~2.34 and an amino $pK_2$ of ~9.69, the pI is calculated as the average: $(2.34 + 9.69) / 2 = \mathbf{6.02}$.
Why other options are incorrect:
  • • 6.6 / 6.8: These values are closer to the pI of other specific amino acids or physiological buffers but do not match the specific titration constants of Alanine.
  • • 7.2: This is near physiological pH and the $pK_a$ of the phosphate buffer system, but it is too high for the isoelectric point of a simple monoamino-monocarboxylic acid like Alanine.
Sulphur containing amino acid is:
  • a) Methionine
  • b) Leucine
  • c) Valine
  • d) Asparagine
Explanation: Methionine is one of the two standard amino acids that contain sulphur (the other being Cysteine). It contains a thioether group in its side chain and serves as the “start” codon (AUG) for protein synthesis.
Why other options are incorrect:
  • • Leucine: A branched-chain amino acid (BCAA) with a purely hydrocarbon side chain; it does not contain sulphur.
  • • Valine: Another branched-chain amino acid with a non-polar hydrocarbon side chain.
  • • Asparagine: A polar amino acid that contains an amide group ($-CONH_2$) in its side chain, but no sulphur.
An example of sulphur containing amino acid is:
  • a) 2-Amino-3-mercaptopropanoic acid
  • b) 2-Amino-3-methylbutanoic acid
  • c) 2-Amino-3-hydroxypropanoic acid
  • d) Amino acetic acid
Explanation: 2-Amino-3-mercaptopropanoic acid is the systematic chemical name for Cysteine. The term “mercapto” indicates the presence of a thiol group (–SH), which contains a sulphur atom.
Why other options are incorrect:
  • • 2-Amino-3-methylbutanoic acid: This is the chemical name for Valine, which contains only carbon, hydrogen, oxygen, and nitrogen.
  • • 2-Amino-3-hydroxypropanoic acid: This is the chemical name for Serine, which has a hydroxyl (–OH) group instead of a sulphur group.
  • • Amino acetic acid: This is the chemical name for Glycine, the simplest amino acid, which has no side chain and thus no sulphur.
All the following are sulphur containing amino acids found in proteins except:
  • a) Cysteine
  • b) Cystine
  • c) Methionine
  • d) Threonine
Explanation: Threonine is an essential amino acid that contains a hydroxyl group (-OH) in its side chain, not sulfur.
Why other options are incorrect:
  • • Cysteine: Contains a highly reactive thiol (-SH) group and is a major sulfur-containing amino acid.
  • • Cystine: Formed when two cysteine molecules are joined by a disulfide bond (-S-S-).
  • • Methionine: An essential amino acid that contains a thioether group and is one of the two primary sulfur-containing amino acids coded by DNA.
An aromatic amino acid is:
  • a) Lysine
  • b) Tyrosine
  • c) Taurine
  • d) Arginine
Explanation: Tyrosine is an aromatic amino acid because it contains a phenyl (benzene) ring with a hydroxyl group attached. The primary aromatic amino acids found in proteins are Phenylalanine, Tyrosine, and Tryptophan.
Why other options are incorrect:
  • • Lysine: A basic amino acid with an aliphatic side chain containing an amino group.
  • • Taurine: A sulfur-containing amino acid derivative that is not used in protein synthesis and does not have an aromatic ring.
  • • Arginine: A basic amino acid containing a complex guanidino group, not an aromatic ring.
The functions of plasma albumin are:
  • a) Osmosis
  • b) Transport
  • c) Immunity
  • d) both (Osmosis) and (Transport)
Explanation: Albumin is the most abundant plasma protein and performs two primary roles: it maintains oncotic (osmotic) pressure to keep fluid in the bloodstream, and it acts as a transport protein for various substances like fatty acids, bilirubin, and drugs.
Why other options are incorrect:
  • • Osmosis: While correct, it is not the only function; albumin also handles transport.
  • • Transport: While correct, it is not the only function; albumin is also vital for osmotic pressure.
  • • Immunity: This is the primary function of globulins (specifically Gamma-globulins or Immunoglobulins), not albumin.
Amino acid with side chain containing basic groups is:
  • a) 2-Amino 5-guanidovaleric acid
  • b) 2-Pyrrolidine carboxylic acid
  • c) 2-Amino 3-mercaptopropanoic acid
  • d) 2-Amino propanoic acid
Explanation: 2-Amino 5-guanidovaleric acid is the IUPAC name for Arginine. It is a basic amino acid because its side chain contains a guanidino group, which is positively charged at physiological pH.
Why other options are incorrect:
  • • 2-Pyrrolidine carboxylic acid: This is Proline, which is an imino acid and does not have a basic side chain.
  • • 2-Amino 3-mercaptopropanoic acid: This is Cysteine, which is a sulfur-containing amino acid with a thiol (-SH) group.
  • • 2-Amino propanoic acid: This is Alanine, which has a simple, non-polar methyl side chain.
An essential amino acid in man is:
  • a) Aspartate
  • b) Tyrosine
  • c) Methionine
  • d) Serine
Explanation: Methionine is an essential amino acid, meaning the human body cannot synthesize it in sufficient quantities and it must be obtained through the diet. It plays a critical role in starting the process of translation (protein synthesis).
Why other options are incorrect:
  • • Aspartate: A non-essential amino acid that the body can produce from oxaloacetate.
  • • Tyrosine: A non-essential amino acid because it can be synthesized in the body from the essential amino acid phenylalanine.
  • • Serine: A non-essential amino acid synthesized from 3-phosphoglycerate (an intermediate of glycolysis).
Non-essential amino acids:
  • a) Are not components of tissue proteins
  • b) Can be synthesized in the body from essential amino acids
  • c) Have no role in the metabolism
  • d) May be synthesized in the body in diseased states
Explanation: Non-essential amino acids are those that the body is capable of synthesizing on its own, either from metabolic intermediates or from essential amino acids (for example, Tyrosine is synthesized from Phenylalanine). Because the body can produce them, they do not strictly need to be obtained from the diet.
Why other options are incorrect:
  • • Are not components of tissue proteins: Incorrect. Both essential and non-essential amino acids are necessary building blocks for tissue proteins.
  • • Have no role in the metabolism: Incorrect. They play vital roles in enzyme function, cell signaling, and energy production.
  • • May be synthesized in the body in diseased states: Incorrect. They are synthesized under normal health conditions. In fact, some non-essential amino acids become “conditionally essential” during disease, meaning the body might fail to produce enough of them.
Which one of the following is semi-essential amino acid for humans?
  • a) Valine
  • b) Arginine
  • c) Lysine
  • d) Tyrosine
Explanation: Arginine (along with Histidine) is considered a semi-essential (or conditionally essential) amino acid. While adults can synthesize enough of it, children and infants cannot produce sufficient quantities to meet the demands of rapid growth, making dietary intake necessary.
Why other options are incorrect:
  • • Valine: An essential amino acid; the body cannot synthesize it at all, regardless of age.
  • • Lysine: An essential amino acid; it must be obtained entirely from the diet.
  • • Tyrosine: A non-essential amino acid; the body can normally synthesize it from the essential amino acid Phenylalanine.
An example of polar amino acid is:
  • a) Alanine
  • b) Leucine
  • c) Arginine
  • d) Valine
Explanation: Arginine is a polar, basic amino acid. Its side chain contains a guanidino group that is hydrophilic and carries a positive charge at physiological pH, making it interact strongly with water.
Why other options are incorrect:
  • • Alanine: A non-polar amino acid with a simple methyl (-CH3) side chain.
  • • Leucine: A non-polar, branched-chain amino acid with a hydrophobic aliphatic side chain.
  • • Valine: Another non-polar, branched-chain amino acid that is hydrophobic and typically found in the interior of proteins.
The amino acid with a nonpolar side chain is:
  • a) Serine
  • b) Valine
  • c) Asparagine
  • d) Threonine
Explanation: Valine is a nonpolar, branched-chain amino acid. Its side chain consists of a hydrophobic isopropyl group, which does not interact with water and is typically buried within the core of protein structures.
Why other options are incorrect:
  • • Serine: A polar amino acid because it contains a hydroxyl (-OH) group in its side chain.
  • • Asparagine: A polar amino acid containing an amide group in its side chain, capable of hydrogen bonding.
  • • Threonine: A polar amino acid because, like serine, it contains a hydroxyl (-OH) group.
A ketogenic amino acid is:
  • a) Valine
  • b) Cysteine
  • c) Leucine
  • d) Threonine
Explanation: Leucine is one of the two purely ketogenic amino acids (the other being Lysine). Its catabolism yields acetyl-CoA or acetoacetate, which are precursors for ketone bodies and cannot be converted into glucose.
Why other options are incorrect:
  • • Valine: A purely glucogenic amino acid; its breakdown leads to the formation of succinyl-CoA, which enters the TCA cycle for gluconeogenesis.
  • • Cysteine: A glucogenic amino acid; it is converted into pyruvate to enter the pathway for glucose synthesis.
  • • Threonine: Primarily a glucogenic amino acid, as it is converted into pyruvate or alpha-ketobutyrate.
An amino acid that does not form an α-helix is:
  • a) Valine
  • b) Proline
  • c) Tyrosine
  • d) Tryptophan
Explanation: Proline is often called a “helix breaker.” Because of its unique cyclic structure (where the side chain is bonded to the nitrogen atom), it creates a rigid kink in the polypeptide chain. Additionally, when Proline is in a peptide bond, it lacks the hydrogen atom on the nitrogen required for the hydrogen bonding that stabilizes an alpha-helix.
Why other options are incorrect:
  • • Valine: While its bulky branched side chain can sometimes favor beta-sheets, it is still capable of being incorporated into alpha-helices.
  • • Tyrosine: An aromatic amino acid that can readily participate in alpha-helical structures.
  • • Tryptophan: A large aromatic amino acid that, despite its size, can be part of an alpha-helix.
An amino acid not found in proteins is:
  • a) β-Alanine
  • b) Proline
  • c) Lysine
  • d) Histidine
Explanation: β-Alanine is a non-proteinogenic amino acid, meaning it is not used in the biosynthesis of proteins. While it is biologically important (it is a component of Pantothenic acid/Vitamin B5 and carnosine), it is not coded for by DNA during translation.
Why other options are incorrect:
  • • Proline: A standard amino acid (technically an imino acid) found in almost all proteins, especially collagen.
  • • Lysine: A standard basic amino acid and an essential building block for proteins.
  • • Histidine: A standard basic amino acid involved in protein structure and the catalytic sites of many enzymes.

Answer Key Summary

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The questions are typically designed to assess the technical skills and knowledge required for the laboratory profession, including the ability to analyze laboratory test results, perform laboratory procedures, and maintain laboratory equipment.

To prepare for these MCQs, candidates should have a thorough understanding of the key concepts and principles of laboratory science. They should also be familiar with common laboratory equipment and procedures, as well as laboratory safety protocols.

Candidates may also benefit from studying specific laboratory science textbooks or taking online courses that cover the material tested in the MCQs. Additionally, practicing sample MCQs and reviewing the answers can help candidates identify areas where they may need to improve their knowledge or skills.

Overall, the MCQs for lab technologists are designed to be challenging and comprehensive, requiring candidates to demonstrate a high level of proficiency in the field of laboratory science.

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