4701 to 4750 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.
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Questions 4701 to 4750
Serum cholesterol is decreased in:
- a) Endemic goitre
- b) Thyrotoxicosis
- c) Myxoedema
- d) Cretinism
Explanation:
In Thyrotoxicosis (hyperthyroidism), thyroid hormones increase the expression of LDL receptors in the liver. This leads to an increased clearance of LDL from the blood, resulting in decreased serum cholesterol levels.
Why other options are incorrect:
Why other options are incorrect:
- • Myxoedema: This is severe hypothyroidism in adults, which typically causes *increased* serum cholesterol due to decreased LDL receptor activity.
- • Cretinism: This is congenital hypothyroidism, which also leads to *increased* cholesterol levels.
- • Endemic goitre: This is usually caused by iodine deficiency; cholesterol levels remain normal unless the condition progresses into hypothyroidism.
The heptose ketose sugar formed as a result of chemical reaction in HMP shunt:
- a) Sedoheptulose
- b) Galactoheptose
- c) Glucoheptose
- d) Mannoheptose
Explanation:
In the non-oxidative phase of the HMP shunt (Hexose Monophosphate Shunt), the enzyme transketolase catalyzes the transfer of a 2-carbon unit to produce Sedoheptulose-7-phosphate, which is a 7-carbon (heptose) ketose sugar.
Why other options are incorrect:
Why other options are incorrect:
- • Galactoheptose: This is not a recognized intermediate in human carbohydrate metabolism or the pentose phosphate pathway.
- • Glucoheptose: While a 7-carbon sugar, it is not an intermediate formed during the HMP shunt.
- • Mannoheptose: This sugar is found in certain plants (like avocados) but is not a product of the HMP shunt in human cells.
The general formula for polysaccharide is:
- a) (C6H12O6)n
- b) (C6H10O5)n
- c) (C6H12O5)n
- d) (C6H19O6)n
Explanation:
Polysaccharides are formed by the condensation of multiple monosaccharide units (like glucose, C₆H₁₂O₆). During the formation of each glycosidic bond, one molecule of water (H₂O) is removed. Therefore, the formula for a hexose-based polysaccharide unit is (C₆H₁₀O₅)n.
Why other options are incorrect:
Why other options are incorrect:
- • (C6H12O6)n: This is the formula for a monosaccharide (like glucose). It doesn’t account for the loss of water needed to link the molecules together.
- • (C6H12O5)n: This formula is mathematically incorrect for standard hexose-based polysaccharides as it suggests only one Hydrogen was removed.
- • (C6H19O6)n: This formula is chemically impossible for a standard carbohydrate, as the Hydrogen count is far too high.
The number of isomers of glucose is:
- a) 4
- b) 8
- c) 12
- d) 16
Explanation:
The number of stereoisomers for a sugar is calculated using the formula 2n, where n is the number of asymmetric (chiral) carbon atoms. Glucose (an aldohexose) has 4 asymmetric carbon atoms (C2, C3, C4, and C5). Therefore, 24 = 16 isomers (8 D-isomers and 8 L-isomers).
Why other options are incorrect:
Why other options are incorrect:
- • 4: This would be the number of isomers for a sugar with only 2 chiral centers (like a triose).
- • 8: This is the number of isomers for an aldopentose (which has 3 chiral centers, 23=8) or the number of D-series isomers of glucose specifically.
- • 12: This number does not fit the 2n mathematical progression for stereoisomers.
The epimer of glucose is:
- a) Fructose
- b) Galactose
- c) Ribose
- d) Deoxyribose
Explanation:
Epimers are stereoisomers that differ in configuration around only one specific carbon atom. Galactose is the 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:
Why other options are incorrect:
- • Fructose: This is a functional isomer of glucose (keto-aldose isomerism), not an epimer.
- • Ribose: This is a pentose (5-carbon sugar), while glucose is a hexose (6-carbon sugar). Epimers must have the same molecular formula.
- • Deoxyribose: This is also a 5-carbon sugar and is missing an oxygen atom at the C-2 position.
The intermediate in hexose monophosphate shunt is:
- a) D-Ribulose
- b) D-Arabinose
- c) D-xylose
- d) D-lyxose
Explanation:
In the oxidative phase of the Hexose Monophosphate (HMP) shunt, 6-phosphogluconate undergoes oxidative decarboxylation to form D-Ribulose-5-phosphate. This is a key 5-carbon ketose intermediate that later isomerizes into Ribose-5-phosphate.
Why other options are incorrect:
Why other options are incorrect:
- • D-Arabinose: This is a pentose sugar found in plant gums and pectins, but it is not a recognized intermediate in the human HMP shunt.
- • D-xylose: While Xylulose-5-phosphate is an intermediate in the non-oxidative phase, D-xylose (the aldose form) is not typically used.
- • D-lyxose: This is a rare pentose sugar that does not play a role in the major metabolic pathways of human carbohydrate metabolism.
Honey contains the hydrolytic product of:
- a) Lactose
- b) Maltose
- c) Inulin
- d) Starch
Explanation:
Honey is primarily composed of Fructose (also known as levulose) and glucose. Inulin is a polysaccharide that, upon hydrolysis, yields fructose. Because fructose is the predominant sugar in honey, it is chemically linked to the hydrolytic products of inulin in biochemical contexts.
Why other options are incorrect:
Why other options are incorrect:
- • Lactose: Found in milk; its hydrolysis yields glucose and galactose, which are not the primary components of honey.
- • Maltose: Hydrolysis yields two glucose units; while honey contains small amounts of maltose, it is not the main identifying component.
- • Starch: A polymer of glucose; its hydrolysis yields maltose and eventually glucose, but not the high fructose levels found in honey.
On boiling, Benedict’s solution is not reduced by:
- a) Sucrose
- b) Lactose
- c) Maltose
- d) Fructose
Explanation:
Sucrose is a non-reducing sugar. It is formed by a glycosidic bond between the anomeric carbon of glucose (C1) and the anomeric carbon of fructose (C2). Because both reducing groups are involved in the linkage, it cannot reduce the cupric ions (Cu²⁺) in Benedict’s solution to cuprous oxide (Cu₂O).
Why other options are incorrect:
Why other options are incorrect:
- • Lactose: A reducing disaccharide (milk sugar) that has a free hemiacetal group on the glucose unit.
- • Maltose: A reducing disaccharide (malt sugar) that has a free hemiacetal group.
- • Fructose: Although a ketose, it is a reducing monosaccharide because it tautomerizes to an aldose in the alkaline environment of Benedict’s reagent.
Glycosides are found in many:
- a) Vitamins
- b) Drugs
- c) Minerals
- d) Nucleoproteins
Explanation:
Glycosides are organic compounds where a sugar is bound to another functional group (the aglycone) via a glycosidic bond. Many important medical drugs are glycosides, such as digitalis (cardiac glycosides like digoxin) used for heart conditions, and various antibiotics or anti-inflammatory agents.
Why other options are incorrect:
Why other options are incorrect:
- • Vitamins: While some vitamins can be glycosylated in specific biological forms, vitamins are primarily classified by their organic function rather than as a major group of glycosides.
- • Minerals: These are inorganic elements (like calcium or iron) and do not contain the organic sugar-linkages that define glycosides.
- • Nucleoproteins: These consist of nucleic acids (DNA/RNA) combined with proteins. While they contain sugars (ribose/deoxyribose), they are not classified as glycosides in the traditional chemical sense.
Galactose on oxidation with conc. HNO3 produces:
- a) Gluconic acid
- b) Saccharic acid
- c) Saccharo Lactone
- d) Mucic acid
Explanation:
When galactose is oxidized with a strong oxidizing agent like concentrated nitric acid (HNO3), both the aldehyde group (C1) and the primary alcoholic group (C6) are converted into carboxylic acid groups. This forms Mucic acid (also known as galactaric acid). Mucic acid is insoluble in water and forms characteristic crystals, which is used as a test for galactose.
Why other options are incorrect:
Why other options are incorrect:
- • Gluconic acid: This is formed by the mild oxidation (using bromine water) of the aldehyde group of glucose, not galactose.
- • Saccharic acid: Also known as glucaric acid, this is the dicarboxylic acid formed when glucose is oxidized with concentrated HNO3.
- • Saccharo Lactone: This is an internal ester (lactone) derived from saccharic acid, not the primary product of galactose oxidation.
The distinguishing test between monosaccharides and disaccharides is:
- a) Bial’s test
- b) Selwanoff’s test
- c) Barfoed’s test
- d) Hydrolysis test
Explanation:
Barfoed’s test is used to distinguish monosaccharides from reducing disaccharides. The reagent is acidic (cupric acetate in acetic acid), and monosaccharides react much faster (usually within 1–2 minutes of boiling) to produce a red precipitate of cuprous oxide, whereas reducing disaccharides take significantly longer (7–12 minutes).
Why other options are incorrect:
Why other options are incorrect:
- • Bial’s test: This test is specifically used to detect pentoses (5-carbon sugars) like ribose.
- • Selwanoff’s test: This is used to distinguish between aldoses and ketoses (e.g., glucose vs. fructose).
- • Hydrolysis test: While hydrolysis breaks disaccharides into monosaccharides, it is a process rather than a specific reagent-based distinguishing color test for identification.
Cellulose is made up of the molecules of:
- a) α-glucose
- b) β-glucose
- c) Both of the above
- d) None of these
Explanation:
Cellulose is a linear, unbranched homopolysaccharide composed of β-D-glucose units. These units are joined by β(1→4) glycosidic bonds. This specific geometry allows cellulose to form long, straight chains that provide structural rigidity to plant cell walls.
Why other options are incorrect:
Why other options are incorrect:
- • α-glucose: These molecules make up Starch (amylose and amylopectin) and Glycogen, linked by α(1→4) and α(1→6) bonds.
- • Both of the above: While both are glucose, biological polymers are highly specific; cellulose exclusively uses the β-isomer.
- • None of these: This is incorrect because β-glucose is the well-established building block of cellulose.
Iodine solution produces no color with:
- a) Cellulose
- b) Starch
- c) Dextrin
- d) Glycogen
Explanation:
The Iodine test depends on the ability of a polysaccharide to form a helical structure that can trap iodine molecules. Cellulose consists of long, straight β-linked chains that do not form these helices; therefore, it does not react with iodine to produce a color change.
Why other options are incorrect:
Why other options are incorrect:
- • Starch: Contains amylose which forms a tight helix that traps iodine to give a characteristic deep blue/black color.
- • Dextrin: These are intermediate products of starch hydrolysis. Higher dextrins (like erythrodextrin) produce a red/brown color with iodine.
- • Glycogen: Highly branched glucose polymer that reacts with iodine to produce a reddish-brown color.
Amylose contains glucose units:
- a) 100–200
- b) 200–300
- c) 300–400
- d) 500–600
Explanation:
Amylose is the linear, non-branched component of starch. In most plants, it typically consists of approximately 200 to 300 D-glucose units (though some sources state it can range from a few hundreds to thousands) linked by α(1→4) glycosidic bonds.
Why other options are incorrect:
Why other options are incorrect:
- • 100–200: This represents a relatively short chain; natural amylose usually exceeds this length to form its stable helical structure.
- • 300–400 / 500–600: While amylose chains can certainly reach these lengths, the “classic” introductory biochemistry range for simple classification questions is usually cited as 200–300 units.
- Note: Amylopectin is much larger, containing 2,000 to 200,000 glucose units, as it is highly branched.
Each branch of amylopectin is at an interval of glucose units:
- a) 14–20
- b) 24–30
- c) 34–40
- d) 44–50
Explanation:
Amylopectin is the branched-chain component of starch. While the main chain consists of α(1→4) linkages, the branches occur via α(1→6) glycosidic bonds. These branch points occur at intervals of approximately every 24 to 30 glucose residues.
Why other options are incorrect:
Why other options are incorrect:
- • 14–20: This branching frequency is higher than that of amylopectin but lower than that of glycogen.
- • 34–40 & 44–50: These ranges describe molecules with very sparse branching, which does not match the structural profile of starch amylopectin.
- Note: In comparison, Glycogen is much more highly branched than amylopectin, with branches occurring every 8 to 12 glucose units.
N-acetylneuraminic acid is an example of:
- a) Sialic acid
- b) Mucic acid
- c) Glucuronic acid
- d) Hippuric acid
Explanation:
N-acetylneuraminic acid (Neu5Ac) is the most common form of Sialic acid found in human cells. It is a 9-carbon acidic sugar typically found at the terminal ends of glycans on cell surfaces and glycoproteins.
Why other options are incorrect:
Why other options are incorrect:
- • Mucic acid: Also known as galactaric acid, it is produced by the nitric acid oxidation of galactose.
- • Glucuronic acid: A sugar acid derived from glucose, used by the body to detoxify substances via glucuronidation.
- • Hippuric acid: A carboxylic acid found in urine, formed by the combination of benzoic acid and glycine; it is not a carbohydrate-based sialic acid.
Blood group substances consist of:
- a) Lactose
- b) Maltose
- c) Fructose
- d) Mucose
Explanation:
Blood group substances (ABH antigens) are complex carbohydrates found on red blood cells and in secretions. These substances are often categorized as mucose (or mucopolysaccharide-like) in nature because they are oligosaccharides attached to proteins or lipids. A key sugar in these structures is L-fucose.
Why other options are incorrect:
Why other options are incorrect:
- • Lactose: A disaccharide found in milk, composed of glucose and galactose.
- • Maltose: A disaccharide composed of two glucose units, usually a product of starch breakdown.
- • Fructose: A simple monosaccharide (fruit sugar) that is not the defining component of blood group specificity.
The component of cartilage and cornea is:
- a) Keratosulphate
- b) Chondroitin sulphate
- c) Cadmium sulphate
- d) Antimony sulphate
Explanation:
Keratosulphate (also known as Keratan sulfate) is a glycosaminoglycan (GAG) that is found primarily in the cornea, cartilage, and bone. In the cornea, it plays a vital role in maintaining tissue hydration and transparency.
Why other options are incorrect:
Why other options are incorrect:
- • Chondroitin sulphate: Although it is the most abundant GAG in cartilage, it is not the primary GAG associated with the specific structural requirements of corneal transparency compared to Keratan sulfate.
- • Cadmium sulphate: An inorganic chemical compound used in vacuum tubes and batteries; it is toxic and not a biological structural component.
- • Antimony sulphate: An inorganic salt used in explosives and pyrotechnics; it has no role in the biological makeup of the eye or joints.
Benedict’s test is less likely to give weakly positive results with concentrated urine due to the action of:
- a) Urea
- b) Uric acid
- c) Ammonium salts
- d) Phosphates
Explanation:
Ammonium salts present in concentrated urine can interfere with Benedict’s test. They react with the copper in the reagent to form a soluble cupric-ammonium complex. This complex prevents the precipitation of cuprous oxide (the red/orange sediment) when only small or trace amounts of reducing sugars are present, thus masking a weakly positive result.
Why other options are incorrect:
Why other options are incorrect:
- • Urea: Does not typically interfere with the reduction of copper ions in Benedict’s reagent.
- • Uric acid: High concentrations of uric acid can actually cause false positive results because uric acid is a weak reducing agent, rather than preventing a positive result.
- • Phosphates: While they may cause a cloudy white precipitate (calcium phosphate) in alkaline conditions, they do not chemically prevent the reduction of copper by sugars.
Active transport of sugar is depressed by the agent:
- a) Oxaloacetate
- b) Fumarate
- c) Malonate
- d) Succinate
Explanation:
Malonate is a potent competitive inhibitor of the enzyme succinate dehydrogenase in the Citric Acid Cycle (Krebs cycle). Active transport of sugars is an energy-dependent process that requires ATP. By inhibiting the respiratory chain and reducing ATP production, malonate depresses the active transport mechanism.
Why other options are incorrect:
Why other options are incorrect:
- • Oxaloacetate: An intermediate in the Krebs cycle; it facilitates the cycle rather than inhibiting it.
- • Fumarate: A metabolic intermediate produced by the oxidation of succinate; it does not inhibit sugar transport.
- • Succinate: A substrate for succinate dehydrogenase; increasing its concentration would actually help overcome malonate inhibition rather than causing depression of transport.
The general test for detection of carbohydrates is:
- a) Iodine test
- b) Molisch test
- c) Barfoed test
- d) Osazone test
Explanation:
The Molisch test is a sensitive chemical test used for the presence of carbohydrates in general. It is based on the dehydration of the carbohydrate by sulfuric acid to produce an aldehyde, which condenses with two molecules of α-naphthol to form a purple or violet-colored ring at the junction of the two liquids.
Why other options are incorrect:
Why other options are incorrect:
- • Iodine test: Specific only for polysaccharides like starch, glycogen, and dextrins; it does not detect simple sugars.
- • Barfoed test: Used specifically to distinguish between monosaccharides and reducing disaccharides.
- • Osazone test: A confirmatory test used to identify specific sugars based on the shape and melting point of crystals formed with phenylhydrazine.
Glucose absorption may be decreased in:
- a) Oedema
- b) Nephritis
- c) Rickets
- d) Osteomalitis
Explanation:
In conditions like Nephritis (inflammation of the kidneys), the renal tubules may be damaged. This impairment reduces the kidney’s ability to reabsorb glucose from the filtrate back into the bloodstream, leading to glucose being excreted in the urine (glycosuria).
Why other options are incorrect:
Why other options are incorrect:
- • Oedema: This is the accumulation of excess fluid in tissues and does not directly interfere with the biochemical mechanism of glucose absorption.
- • Rickets: This is a skeletal disorder caused by Vitamin D deficiency affecting calcium and phosphate metabolism, not glucose absorption.
- • Osteomalitis: (Osteomyelitis) This is an infection or inflammation of the bone or bone marrow and has no direct impact on glucose reabsorption.
Glycogen synthetase activity is depressed by:
- a) Glucose
- b) Insulin
- c) Cyclic AMP
- d) Fructokinase
Explanation:
Cyclic AMP (cAMP) activates Protein Kinase A, which phosphorylates the active form of glycogen synthase (Type I or a) into its inactive form (Type D or b). Therefore, an increase in cAMP levels leads to the depression of glycogen synthesis.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: High levels of glucose actually stimulate glycogen synthesis to store energy.
- • Insulin: Insulin promotes glycogen synthesis by activating phosphatases that dephosphorylate (activate) glycogen synthase.
- • Fructokinase: This is an enzyme involved in the metabolism of fructose, not a direct regulator of the glycogen synthase enzyme.
The branching enzyme acts on the glycogen when the glycogen chain has been lengthened to between glucose units:
- a) 1 and 6
- b) 2 and 7
- c) 3 and 9
- d) 6 and 11
Explanation:
The branching enzyme (amylo-α(1→4)→α(1→6)transglucosidase) acts on a glycogen chain once it has reached a minimum length of about 11 glucose units. It then transfers a fragment of 6 to 7 units to a more interior site to create an α-1,6-glycosidic bond.
Why other options are incorrect:
Why other options are incorrect:
- • 1 and 6: This range is far too short; the enzyme requires a longer chain to maintain structural stability before creating a branch.
- • 2 and 7: At this length, the chain is still being elongated by glycogen synthase and has not reached the threshold for branching.
- • 3 and 9: While closer, the biochemical requirement for the branching enzyme is generally a chain of at least 11 residues.
Cyclic AMP is formed from ATP by the enzyme adenylate cyclase which is activated by the hormone:
- a) Insulin
- b) Epinephrine
- c) Testosterone
- d) Progesterone
Explanation:
Epinephrine (and Glucagon) binds to specific cell surface receptors (G-protein coupled receptors) which activate adenylate cyclase. This enzyme then catalyzes the conversion of ATP into Cyclic AMP (cAMP), which acts as a second messenger to trigger glycogen breakdown.
Why other options are incorrect:
Why other options are incorrect:
- • Insulin: Insulin generally has the opposite effect; it lowers cAMP levels by activating phosphodiesterase, an enzyme that breaks cAMP down.
- • Testosterone: This is a steroid hormone that primarily acts through intracellular nuclear receptors to regulate gene expression rather than the cAMP pathway.
- • Progesterone: Like testosterone, this is a steroid hormone that mainly influences the body through nuclear receptors and transcription factors.
Hexokinase has a high affinity for glucose than:
- a) Fructokinase
- b) Galactokinase
- c) Glucokinase
- d) All of the above
Explanation:
Hexokinase has a very low Km (Michaelis constant), which means it has a high affinity for glucose, allowing it to function even at low blood sugar concentrations. In contrast, Glucokinase (Hexokinase IV) has a high Km and thus a low affinity, only becoming fully active when glucose levels are high.
Why other options are incorrect:
Why other options are incorrect:
- • Fructokinase: This enzyme is specific for the phosphorylation of fructose and does not primarily compete with Hexokinase for glucose binding.
- • Galactokinase: This enzyme is specific for galactose. While it is a kinase, it is not compared to Hexokinase in terms of glucose affinity.
- • All of the above: While Hexokinase is the most efficient glucose-phosphorylating enzyme among these, the specific biochemical distinction regarding “affinity” is traditionally made between Hexokinase and Glucokinase.
Dihydroxyacetone phosphate and glyceraldehyde-3-phosphate are interconverted by:
- a) Triose isomerase
- b) Phosphotriose isomerase
- c) Diphosphotriose isomerase
- d) Dihydroxyacetone phosphorylase
Explanation:
Phosphotriose isomerase (also known as Triose Phosphate Isomerase or TPI) is the enzyme that catalyzes the reversible interconversion of the triose phosphate isomers dihydroxyacetone phosphate (DHAP) and D-glyceraldehyde-3-phosphate. This is a crucial step in glycolysis to ensure all carbons from glucose can be processed.
Why other options are incorrect:
Why other options are incorrect:
- • Triose isomerase: While similar, the standard biochemical name for the enzyme specifically acting on these phosphorylated sugars is Phosphotriose isomerase.
- • Diphosphotriose isomerase: DHAP and G3P are monophosphates (containing only one phosphate group), so a “diphospho” enzyme would not be applicable here.
- • Dihydroxyacetone phosphorylase: Phosphorylases are enzymes that introduce a phosphate group into an organic molecule; they do not perform isomerization (changing the structure of the molecule).
Citrate is converted to isocitrate by aconitase which contains:
- a) Ca++
- b) Fe++
- c) Zn++
- d) Mg++
Explanation:
Aconitase is an iron-sulfur protein. It contains an iron-sulfur cluster [4Fe-4S] where the Fe++ (ferrous iron) ion directly participates in the binding of the substrate (citrate) and facilitates the dehydration/rehydration reaction.
Why other options are incorrect:
Why other options are incorrect:
- • Ca++: Calcium ions are often involved in signal transduction and muscle contraction, but they do not serve as the cofactor for aconitase.
- • Zn++: Zinc is a common cofactor for enzymes like carbonic anhydrase and alcohol dehydrogenase, but it is not part of the aconitase active site.
- • Mg++: Magnesium is frequently required by enzymes that utilize ATP (like kinases) to stabilize the phosphate groups, but aconitase specifically requires iron.
The reaction succinyl COA to succinate requires:
- a) CDP
- b) ADP
- c) GDP
- d) NADP+
Explanation:
The conversion of succinyl-CoA to succinate is catalyzed by the enzyme Succinyl-CoA synthetase (also known as succinic thiokinase). In animal tissues, this reaction is a substrate-level phosphorylation that specifically requires GDP (Guanosine diphosphate) and inorganic phosphate to form GTP.
Why other options are incorrect:
Why other options are incorrect:
- • CDP: Cytidine diphosphate is primarily involved in lipid metabolism (like the synthesis of phospholipids) and is not used in the Citric Acid Cycle.
- • ADP: While some bacterial and plant isoforms of this enzyme use ADP, the standard biochemical pathway in human/animal metabolism highlights the use of GDP.
- • NADP+: This is a coenzyme used in redox reactions (especially in the Pentose Phosphate Pathway and fatty acid synthesis), not in substrate-level phosphorylation.
The carrier of the citric acid cycle is:
- a) Succinate
- b) Fumarate
- c) Malate
- d) Oxaloacetate
Explanation:
Oxaloacetate is considered the “carrier” of the Citric Acid Cycle (Krebs cycle). It reacts with Acetyl-CoA to form Citrate, initiating the cycle. Since Oxaloacetate is regenerated at the end of the cycle, it acts as a catalytic carrier that allows the oxidation of multiple acetyl units.
Why other options are incorrect:
Why other options are incorrect:
- • Succinate: This is a 4-carbon intermediate produced later in the cycle, but it does not react with Acetyl-CoA to start the process.
- • Fumarate: An intermediate formed by the oxidation of succinate; it is a middle step in the regeneration of oxaloacetate.
- • Malate: The intermediate immediately preceding oxaloacetate; while crucial, it is not the molecule that accepts the entry of the acetyl group.
UDPG is oxidized to UDP glucuronic acid by UDP dehydrogenase in presence of:
- a) FAD+
- b) NAD+
- c) NADP+
- d) ADP+
Explanation:
The enzyme UDP-glucose dehydrogenase catalyzes the two-step oxidation of UDP-glucose to UDP-glucuronic acid. This reaction requires NAD+ as the electron acceptor. Specifically, two molecules of NAD+ are reduced to NADH during the conversion of the primary alcohol group of glucose into a carboxyl group.
Why other options are incorrect:
Why other options are incorrect:
- • FAD+: Flavin Adenine Dinucleotide is typically used in reactions involving the formation of double bonds (like in the Citric Acid Cycle), not for this specific carbohydrate oxidation.
- • NADP+: While structurally similar to NAD+, NADP+ is generally reserved for reductive biosynthetic pathways (like fatty acid synthesis) and the Pentose Phosphate Pathway.
- • ADP+: This is not a standard biological redox cofactor. ADP (Adenosine diphosphate) is involved in energy transfer, and the “plus” notation is not biologically applicable here.
Galactose is phosphorylated by galactokinase to form:
- a) Galactose-6-phosphate
- b) Galactose-1, 6 diphosphate
- c) Galactose-1-phosphate
- d) All of these
Explanation:
In the Leloir pathway (the main pathway for galactose metabolism), the enzyme galactokinase catalyzes the phosphorylation of galactose using ATP. Unlike glucose (which is usually phosphorylated at the C6 position), galactose is specifically phosphorylated at the C1 position to form Galactose-1-phosphate.
Why other options are incorrect:
Why other options are incorrect:
- • Galactose-6-phosphate: While Hexokinase can occasionally phosphorylate galactose at C6, the specific enzyme galactokinase only produces the C1 variant.
- • Galactose-1, 6 diphosphate: This is a rare intermediate and is not the direct product of the initial phosphorylation of galactose by galactokinase.
- • All of these: Galactokinase is highly specific for the production of Galactose-1-phosphate.
The conversion of alanine to glucose is termed:
- a) Glycolysis
- b) Oxidative decarboxylation
- c) Specific dynamic action
- d) Gluconeogenesis
Explanation:
Gluconeogenesis is the metabolic pathway that results in the generation of glucose from non-carbohydrate carbon substrates. Alanine is a glucogenic amino acid; it is converted into pyruvate in the liver, which then serves as a precursor for glucose synthesis.
Why other options are incorrect:
Why other options are incorrect:
- • Glycolysis: This is the metabolic pathway that breaks down glucose into pyruvate to produce energy, which is the reverse of the process described.
- • Oxidative decarboxylation: This refers to a chemical reaction in which a carboxyl group is removed, forming carbon dioxide (e.g., the conversion of pyruvate to acetyl-CoA).
- • Specific dynamic action (SDA): This refers to the energy expenditure (heat produced) by the body during the processing of food for use and storage.
The blood sugar raising action of the hormones of suprarenal cortex is due to:
- a) Gluconeogenesis
- b) Glycogenolysis
- c) Glucagon-like activity
- d) Due to inhibition of glomerular filtration
Explanation:
The suprarenal (adrenal) cortex secretes glucocorticoids, such as cortisol. These hormones increase blood glucose levels by promoting gluconeogenesis (the synthesis of glucose from non-carbohydrate sources like amino acids and glycerol) and by decreasing the peripheral utilization of glucose.
Why other options are incorrect:
Why other options are incorrect:
- • Glycogenolysis: This is the primary mechanism of hormones from the adrenal medulla (Epinephrine) and the pancreas (Glucagon), rather than the cortex.
- • Glucagon-like activity: While the end result (high blood sugar) is similar, the molecular pathway of cortical hormones involves protein catabolism, which is distinct from glucagon’s mechanism.
- • Inhibition of glomerular filtration: This would affect the excretion of substances in urine but is not a metabolic mechanism for raising blood glucose levels.
Under anaerobic conditions the glycolysis of one mole of glucose yields __ moles of ATP:
- a) One
- b) Two
- c) Eight
- d) Thirty
Explanation:
In anaerobic glycolysis, glucose is converted to lactate. During this process, 2 ATP molecules are consumed in the preparatory phase and 4 ATP molecules are produced in the payoff phase, resulting in a net yield of 2 ATP per mole of glucose.
Why other options are incorrect:
Why other options are incorrect:
- • One: This is incorrect as the net production is always 2 ATP per glucose molecule entering the pathway.
- • Eight: This value is sometimes used to represent the net yield of aerobic glycolysis (2 ATP + 2 NADH yielding roughly 6 ATP), but not anaerobic.
- • Thirty: This represents the approximate total yield of ATP from the complete aerobic oxidation of glucose (including the Krebs Cycle and Electron Transport Chain).
Which of the following metabolite integrates glucose and fatty acid metabolism?
- a) Acetyl CoA
- b) Pyruvate
- c) Citrate
- d) Lactate
Explanation:
Acetyl CoA is the central “crossroads” molecule of metabolism. It is the end product of both Glycolysis (via pyruvate) and Fatty Acid Oxidation ($\beta$-oxidation). It enters the Citric Acid Cycle to produce energy and also serves as the starting substrate for fatty acid synthesis.
Why other options are incorrect:
Why other options are incorrect:
- • Pyruvate: While a key intermediate in glucose metabolism, fatty acids are not converted back into pyruvate in animal tissues.
- • Citrate: This is an intermediate of the Citric Acid Cycle. While it helps transport Acetyl CoA out of the mitochondria for fat synthesis, it isn’t the primary shared breakdown product.
- • Lactate: This is a product of anaerobic glucose metabolism and is not directly involved in the breakdown or synthesis of fatty acids.
Cerebrosides consist mostly of this sugar:
- a) Glucose
- b) Fructose
- c) Galactose
- d) Arabinose
Explanation:
Cerebrosides are a type of glycosphingolipid. While there are both glucocerebrosides and galactocerebrosides, Galactose is the characteristic sugar found in the cerebrosides of the brain and nervous system (specifically in the myelin sheath).
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: Although glucocerebrosides do exist (often found in non-neural tissues), galactose is considered the primary sugar when discussing classical cerebrosides.
- • Fructose: This is a keto-sugar primarily involved in energy metabolism and is not a structural component of sphingolipids.
- • Arabinose: This is a five-carbon plant sugar (pentose) and is not found in the structure of mammalian cerebrosides.
Glucose will be converted into fatty acids if the diet has excess of:
- a) Carbohydrates
- b) Proteins
- c) Fat
- d) Vitamins
Explanation:
When Carbohydrates are consumed in excess of the body’s immediate energy needs and glycogen storage capacity, the extra glucose is processed through glycolysis to form Acetyl-CoA. This Acetyl-CoA is then used as a building block for de novo lipogenesis (fatty acid synthesis) in the liver and adipose tissue.
Why other options are incorrect:
Why other options are incorrect:
- • Proteins: While excess protein can eventually be converted to fat, the specific pathway of converting glucose to fatty acids is triggered by a surplus of dietary carbohydrates.
- • Fat: If the diet is high in fat, the body stores the dietary fat directly. It does not need to convert glucose into fatty acids because energy is already available in lipid form.
- • Vitamins: Vitamins are micronutrients that do not provide the carbon chains or energy required to synthesize fatty acids.
The purple ring of Molisch reaction is due to
- a) Furfural
- b) Furfural + α-Naphthol
- c) α-Naphthol
- d) Furfurol + H₂SO₄ + α-Naphthol
Explanation:
In the Molisch reaction, carbohydrates are dehydrated by concentrated H₂SO₄ to form furfural or hydroxymethylfurfural. These compounds react with α-naphthol to produce the characteristic purple/violet ring.
Why other options are incorrect:
Why other options are incorrect:
- • Furfural: Furfural alone does not produce the purple ring.
- • α-Naphthol: It acts as the reagent but does not form the colored ring by itself.
- • Furfurol + H₂SO₄ + α-Naphthol: This is not the standard description of the color-forming reaction.
One of the following enzymes does not change glycogen synthase ‘a’ to ‘b’:
- a) Glycogen synthase kinases 3, 4, 5
- b) Ca2+ calmodulin phosphorylase kinase
- c) Ca2+ calmodulin dependent protein kinase
- d) Glycogen phosphorylase a
Explanation:
The conversion of glycogen synthase from its active form (a) to its inactive form (b) is a process of phosphorylation. This is carried out by various kinases. Glycogen phosphorylase a is an enzyme involved in the breakdown of glycogen; it is not a kinase and does not have the ability to phosphorylate or regulate the activity of glycogen synthase.
Why other options are incorrect:
Why other options are incorrect:
- • GSK 3, 4, 5: These are specific kinases whose primary role is to phosphorylate glycogen synthase to inhibit its activity.
- • Ca2+ calmodulin phosphorylase kinase: This is a multi-functional kinase that can phosphorylate both glycogen phosphorylase (to activate it) and glycogen synthase (to inactivate it).
- • Ca2+ calmodulin dependent protein kinase: This enzyme responds to increased calcium levels and acts as a kinase to phosphorylate and inactivate glycogen synthase.
In the EM (Embden-Meyerhof) pathway, 2-phosphoglycerate is converted to:
- a) Phospho enol pyruvate
- b) Enol pyruvate
- c) Di hydroxy acetone phosphate (DHAP)
- d) 1,3 bisphosphoglycerate
Explanation:
In the ninth step of glycolysis (EM pathway), the enzyme Enolase catalyzes the dehydration of 2-phosphoglycerate to form Phosphoenolpyruvate (PEP). This reaction introduces a high-energy phosphate bond in preparation for the subsequent production of ATP.
Why other options are incorrect:
Why other options are incorrect:
- • Enol pyruvate: This is a transient, unstable form of pyruvate that exists briefly after the PEP is dephosphorylated, but it is not the direct product of 2-phosphoglycerate.
- • DHAP: This triose phosphate is formed much earlier in the pathway during the cleavage of Fructose-1,6-bisphosphate.
- • 1,3 bisphosphoglycerate: This is an intermediate formed earlier in the payoff phase by the oxidation of Glyceraldehyde-3-phosphate.
An anaplerotic reaction which sustains the availability of oxaloacetate is the carboxylation of:
- a) Glutamate
- b) Pyruvate
- c) Citrate
- d) Succinate
Explanation:
Anaplerotic reactions are “filling up” reactions that replenish Citric Acid Cycle intermediates. The most important anaplerotic reaction in animal tissues is the carboxylation of pyruvate to form oxaloacetate, catalyzed by the enzyme pyruvate carboxylase. This ensures that oxaloacetate levels remain sufficient to keep the cycle running.
Why other options are incorrect:
Why other options are incorrect:
- • Glutamate: While glutamate can replenish the cycle by being converted to $\alpha$-ketoglutarate, this is typically via transamination or oxidative deamination, not carboxylation to oxaloacetate.
- • Citrate: Citrate is the product formed when oxaloacetate reacts with acetyl-CoA; it is not the precursor used in an anaplerotic carboxylation reaction.
- • Succinate: Succinate is an intermediate within the cycle itself. Its conversion to oxaloacetate is a standard part of the cycle’s pathway, not an external “filling up” carboxylation reaction.
Specific test for ketohexoses:
- a) Seliwanoff’s test
- b) Osazone test
- c) Molisch test
- d) None of these
Explanation:
Seliwanoff’s test is used to distinguish between aldoses and ketoses. Ketohexoses (like fructose) react with resorcinol in the presence of concentrated HCl to produce a cherry-red color much faster than aldoses do.
Why other options are incorrect:
Why other options are incorrect:
- • Osazone test: This is a general test for reducing sugars (both aldoses and ketoses) based on the formation of characteristic crystals with phenylhydrazine.
- • Molisch test: This is a general screening test for all carbohydrates and does not distinguish between different types of sugars.
- • None of these: This is incorrect because Seliwanoff’s test is specifically designed for this purpose.
Two important byproducts of HMP shunt are:
- a) NADH and pentose sugars
- b) NADPH and pentose sugars
- c) Pentose sugars and 4 membered sugars
- d) Pentose sugars and sedoheptulose
Explanation:
The HMP Shunt (Hexose Monophosphate Shunt), also known as the Pentose Phosphate Pathway, has two primary functions: generating NADPH for reductive biosynthetic reactions (like fatty acid synthesis) and producing Pentose sugars (specifically Ribose-5-phosphate) for the synthesis of nucleotides and nucleic acids.
Why other options are incorrect:
Why other options are incorrect:
- • NADH: NADH is primarily produced in glycolysis and the Citric Acid Cycle for ATP production. The HMP shunt specifically produces NADPH, which has a different metabolic role.
- • 4 membered sugars: While Erythrose-4-phosphate is an intermediate in the non-oxidative phase, it is not considered one of the two “main” primary products of the pathway.
- • Sedoheptulose: Sedoheptulose-7-phosphate is a 7-carbon intermediate of the non-oxidative phase, but it is typically recycled back into glycolytic intermediates rather than being a final “byproduct” of interest.
Pyruvate dehydrogenase complex and α-ketoglutarate dehydrogenase complex require the following for their oxidative decarboxylation:
- a) COASH and Lipoic acid
- b) NAD+ and FAD
- c) COASH and TPP
- d) COASH, TPP, NAD+, FAD, Lipoate
Explanation:
Both the Pyruvate Dehydrogenase (PDH) and α-ketoglutarate dehydrogenase complexes are multi-enzyme systems that require five coenzymes to function:
1. TPP (Thiamine pyrophosphate)
2. Lipoate (Lipoic acid)
3. CoA-SH (Coenzyme A)
4. FAD (Flavin adenine dinucleotide)
5. NAD+ (Nicotinamide adenine dinucleotide)
Why other options are incorrect:
Why other options are incorrect:
- • Options a, b, and c: While these options list real requirements, they are incomplete. All five cofactors must be present simultaneously for the oxidative decarboxylation process to occur.
The four membered aldose sugar phosphate formed in HMP shunt pathway is:
- a) Xylulose P
- b) Erythrulose P
- c) Erythrose P
- d) Ribulose P
Explanation:
Erythrose-4-phosphate (Erythrose P) is a four-carbon (tetrose) aldose sugar phosphate produced during the non-oxidative phase of the Hexose Monophosphate (HMP) shunt. It is formed by the action of the enzyme transaldolase and serves as a precursor for the synthesis of aromatic amino acids.
Why other options are incorrect:
Why other options are incorrect:
- • Xylulose P: This is a five-carbon sugar (pentose) and it is a ketose, not an aldose.
- • Erythrulose P: While this is a four-carbon sugar, it is a ketose. Erythrose is the aldose form.
- • Ribulose P: This is a five-carbon sugar (pentose) and it is a ketose formed in the oxidative phase of the shunt.
Cane sugar (Sucrose) injected into blood is:
- a) changed to fructose
- b) changed to glucose
- c) undergoes no significant change
- d) changed to glucose and fructose
Explanation:
The enzyme sucrase (which breaks down sucrose into glucose and fructose) is located in the intestinal brush border, not in the blood or tissues. Therefore, if sucrose is injected directly into the bloodstream, the body cannot metabolize it, and it undergoes no significant change before being excreted unchanged in the urine.
Why other options are incorrect:
Why other options are incorrect:
- • changed to glucose/fructose: These conversions only happen in the digestive tract where sucrase is present. The blood lacks the necessary hydrolase enzymes for this reaction.
- • changed to glucose and fructose: While this is what happens when you eat sugar, it does not happen when sugar is injected intravenously.
Pentose production is increased in:
- a) HMP shunt
- b) Uronic acid pathway
- c) EM pathway
- d) TCA cycle
Explanation:
The HMP shunt (Hexose Monophosphate Shunt), also known as the Pentose Phosphate Pathway, is the major metabolic pathway responsible for the production of pentose sugars, specifically Ribose-5-phosphate, which is essential for the synthesis of RNA, DNA, and nucleotide coenzymes.
Why other options are incorrect:
Why other options are incorrect:
- • Uronic acid pathway: While this pathway involves some pentoses (like xylulose), its primary purpose is the production of glucuronic acid for detoxification and proteoglycan synthesis.
- • EM pathway: (Glycolysis) This pathway breaks down glucose into pyruvate to produce ATP; it does not produce pentoses.
- • TCA cycle: (Krebs Cycle) This is the final common pathway for the oxidation of fuel molecules to CO2 and water; it does not involve pentose synthesis.
Conversion of Alanine to carbohydrate is termed:
- a) Glycogenesis
- b) Gluconeogenesis
- c) Glycogenolysis
- d) Photosynthesis
Explanation:
Gluconeogenesis is the metabolic process by which the body produces glucose from non-carbohydrate precursors. Alanine is a glucogenic amino acid; it is converted into pyruvate in the liver, which then enters the gluconeogenic pathway to become glucose.
Why other options are incorrect:
Why other options are incorrect:
- • Glycogenesis: This is the process of synthesizing glycogen from glucose molecules for storage.
- • Glycogenolysis: This is the biochemical breakdown of glycogen back into glucose.
- • Photosynthesis: This is the process used by plants and other organisms to convert light energy into chemical energy (glucose) from carbon dioxide and water.
The following is an enzyme required for glycolysis:
- a) Pyruvate kinase
- b) Pyruvate carboxylase
- c) Glucose-6-phosphatase
- d) Glycerokinase
Explanation:
Pyruvate kinase is the enzyme that catalyzes the final step of glycolysis. it transfers a phosphate group from phosphoenolpyruvate (PEP) to ADP, yielding one molecule of pyruvate and one molecule of ATP. This is one of the three irreversible, regulatory steps of the pathway.
Why other options are incorrect:
Why other options are incorrect:
- • Pyruvate carboxylase: This is a key enzyme in gluconeogenesis (the synthesis of glucose), not glycolysis. It converts pyruvate into oxaloacetate.
- • Glucose-6-phosphatase: This enzyme is involved in gluconeogenesis and glycogenolysis. It removes the phosphate from glucose-6-phosphate to release free glucose into the blood.
- • Glycerokinase: This enzyme is involved in the metabolism of glycerol (lipids), phosphorylating glycerol to glycerol-3-phosphate. It is not part of the main glycolytic pathway.
Answer Key Summary
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.
📚 This article may have been prepared using trusted medical literature, clinical guidelines, textbooks, and peer-reviewed sources.
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