4601 to 4650 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 4601 to 4650
The most important epimer of glucose is:
- a) Galactose
- b) Fructose
- c) Arabinose
- d) Xylose
Explanation:
Epimers are diastereomers that differ in configuration at only one specific chiral carbon atom. Galactose is the C-4 epimer of glucose, meaning they are identical except for the orientation of the -OH group at the fourth carbon.
Why other options are incorrect:
Why other options are incorrect:
- • Fructose: It is a functional isomer (keto-sugar) of glucose, not an epimer.
- • Arabinose: This is a five-carbon sugar (pentose), whereas glucose is a six-carbon sugar (hexose). Epimers must have the same number of carbons.
- • Xylose: Similar to arabinose, xylose is a pentose and cannot be an epimer of the hexose glucose.
α-D-glucose and β-D-glucose are:
- a) Stereoisomers
- b) Epimers
- c) Anomers
- d) Keto-aldo pairs
Explanation:
α-D-glucose and β-D-glucose are Anomers. They differ only in the configuration of the hydroxyl (-OH) group at the C-1 carbon atom (the anomeric carbon) when the sugar forms a ring structure.
Why other options are incorrect:
Why other options are incorrect:
- • Stereoisomers: While anomers are a type of stereoisomer, “Anomers” is the specific and most accurate term for this relationship.
- • Epimers: These are sugars that differ at any one chiral center other than the anomeric carbon (e.g., Glucose and Galactose at C-4).
- • Keto-aldo pairs: This refers to functional isomers like Glucose (an aldose) and Fructose (a ketose).
α-D-glucose (+112.2°) → +52.7° ← +18.7° β-D-glucose. The process described above for glucose represents:
- a) Optical isomerism
- b) Mutarotation
- c) Epimerisation
- d) D and L isomerism
Explanation:
The phenomenon of a change in the specific optical rotation of an optically active compound (like α or β glucose) until it reaches a constant equilibrium value is called Mutarotation. In water, α-D-glucose (+112.2°) and β-D-glucose (+18.7°) both slowly convert into an equilibrium mixture with a rotation of +52.7°.
Why other options are incorrect:
Why other options are incorrect:
- • Optical isomerism: This is a general category of isomerism; mutarotation is the specific process of interconverting these isomers.
- • Epimerisation: This involves the conversion of one epimer to another (changing configuration at a non-anomeric carbon), whereas this is a change at the anomeric carbon.
- • D and L isomerism: This refers to the configuration of the molecule relative to glyceraldehyde and does not describe the dynamic change in rotation in solution.
Compounds having the same structural formula but differing in spatial configuration are known as:
- a) Stereoisomers
- b) Anomers
- c) Optical isomers
- d) Epimers
Explanation:
Stereoisomers is the broad, general term for molecules that have the same molecular formula and sequence of bonded atoms (structure), but differ in the three-dimensional orientations of their atoms in space.
Why other options are incorrect:
Why other options are incorrect:
- • Anomers: These are a specific sub-type of stereoisomer that differ only at the anomeric carbon (C-1 for aldoses).
- • Optical isomers: This is a sub-type of stereoisomer that differs in its ability to rotate plane-polarized light. While correct in some cases, “Stereoisomers” is the more complete definition for “spatial configuration” generally.
- • Epimers: These are a specific sub-type of diastereomers (stereoisomers) that differ in configuration at only one chiral center.
In glucose, the orientation of the —H and —OH groups around the carbon atom 5 (adjacent to the terminal primary alcohol carbon) determines:
- a) D or L series
- b) Dextro or levorotatory
- c) α and β anomers
- d) Epimers
Explanation:
The D or L series is determined by the configuration of the chiral carbon furthest from the carbonyl group (C1). In glucose, this is Carbon-5 (C5). If the —OH group on C5 is on the right (in a Fischer projection), it is the D-series; if it is on the left, it is the L-series.
Why other options are incorrect:
Why other options are incorrect:
- • Dextro or levorotatory: This refers to the actual direction (+ or -) in which a substance rotates plane-polarized light, which must be measured experimentally and is not always determined by the D/L configuration.
- • α and β anomers: This is determined by the orientation of the —OH group around the anomeric carbon (C1), not C5.
- • Epimers: Epimers are sugars that differ at any single chiral center. While C5 could be a point of difference, the standard convention for C5 specifically defines the D/L series.
The carbohydrate of the blood group substances is:
- a) Sucrose
- b) Fucose
- c) Arabinose
- d) Maltose
Explanation:
L-Fucose (a deoxyhexose sugar) is a key carbohydrate component found in the oligosaccharide chains of blood group substances (antigens) located on the surface of red blood cells. It is especially critical as the terminal sugar for the H-antigen.
Why other options are incorrect:
Why other options are incorrect:
- • Sucrose: This is common table sugar (glucose + fructose) and is not a component of cell surface antigens.
- • Arabinose: This is a five-carbon sugar (pentose) primarily found in plant hemicellulose and bacterial cell walls.
- • Maltose: This is a disaccharide formed from two glucose units, typically resulting from the breakdown of starch, and does not form part of blood group structures.
Erythromycin contains:
- a) Dimethyl amino sugar
- b) Trimethyl amino sugar
- c) Sterol and sugar
- d) Glycerol and sugar
Explanation:
Erythromycin is a macrolide antibiotic that consists of a large lactone ring attached to two sugar moieties: L-cladinose and D-desosamine. Desosamine is specifically a dimethyl amino sugar.
Why other options are incorrect:
Why other options are incorrect:
- • Trimethyl amino sugar: The amino group in desosamine is substituted with two methyl groups (dimethyl), not three.
- • Sterol and sugar: This combination is characteristic of cardiac glycosides (like digitalis), not macrolide antibiotics.
- • Glycerol and sugar: This describes certain glycolipids or glycerides; erythromycin has a complex macrocyclic lactone ring instead of a glycerol backbone.
A sugar alcohol is:
- a) Mannitol
- b) Trehalose
- c) Xylulose
- d) Arabinose
Explanation:
Mannitol is a sugar alcohol (also known as a polyol). It is formed by the reduction of the aldehyde group of mannose into a hydroxyl group. It is commonly used as an osmotic diuretic.
Why other options are incorrect:
Why other options are incorrect:
- • Trehalose: This is a non-reducing disaccharide composed of two glucose units, not an alcohol.
- • Xylulose: This is a ketose sugar (a pentose), specifically a functional isomer, not a sugar alcohol.
- • Arabinose: This is an aldose sugar (a pentose) found in plant polysaccharides.
The major sugar of insect hemolymph is:
- a) Glycogen
- b) Pectin
- c) Trehalose
- d) Sucrose
Explanation:
Trehalose is a non-reducing disaccharide composed of two glucose units. It is the primary circulating sugar in the hemolymph (blood) of most insects, serving as a highly efficient and stable energy source for flight and metabolic activities.
Why other options are incorrect:
Why other options are incorrect:
- • Glycogen: This is a storage polysaccharide found in tissues (like the fat body or muscle) rather than being the main circulating sugar in the blood.
- • Pectin: This is a structural polysaccharide found in the cell walls of terrestrial plants, not in animal fluids.
- • Sucrose: While insects may ingest sucrose from plant nectar, it is typically broken down and converted into trehalose for circulation in the hemolymph.
The sugar found in DNA is:
- a) Xylose
- b) Ribose
- c) Deoxyribose
- d) Ribulose
Explanation:
DNA stands for Deoxyribonucleic Acid. Its backbone is composed of a five-carbon sugar called 2-deoxy-D-ribose. It differs from ribose by the absence of an oxygen atom at the C-2 position.
Why other options are incorrect:
Why other options are incorrect:
- • Xylose: A pentose sugar found in plant materials like wood, but not involved in the structure of nucleic acids.
- • Ribose: This is the pentose sugar found in RNA (Ribonucleic Acid), not DNA.
- • Ribulose: A ketopentose sugar that acts as an intermediate in the pentose phosphate pathway and is crucial in photosynthesis, but not a component of DNA.
The sugar found in RNA is:
- a) Ribose
- b) Deoxyribose
- c) Ribulose
- d) Erythrose
Explanation:
RNA (Ribonucleic Acid) contains Ribose, which is a pentose sugar (five-carbon sugar). Unlike deoxyribose, ribose has a hydroxyl (-OH) group attached to its second carbon (C2) atom.
Why other options are incorrect:
Why other options are incorrect:
- • Deoxyribose: This sugar is specifically found in DNA (Deoxyribonucleic Acid). It lacks one oxygen atom at the C2 position compared to ribose.
- • Ribulose: This is a ketose sugar that serves as an intermediate in the pentose phosphate pathway and photosynthesis (as RuBP), but it is not part of the RNA backbone.
- • Erythrose: This is a four-carbon tetrose sugar, not a five-carbon sugar, and does not occur in nucleic acids.
The sugar found in milk is:
- a) Galactose
- b) Glucose
- c) Fructose
- d) Lactose
Explanation:
Lactose is commonly known as “milk sugar.” It is a disaccharide composed of one molecule of D-galactose and one molecule of D-glucose joined by a β(1→4) glycosidic linkage. It makes up around 2–8% of milk by weight.
Why other options are incorrect:
Why other options are incorrect:
- • Galactose: While it is a component of lactose, it is rarely found in its free (monosaccharide) form in milk.
- • Glucose: Like galactose, it is a building block of lactose but is not the specific sugar identified as milk sugar.
- • Fructose: Known as fruit sugar, it is primarily found in fruits, honey, and root vegetables, not in animal milk.
Invert sugar is:
- a) Lactose
- b) Sucrose
- c) Hydrolytic products of sucrose
- d) Fructose
Explanation:
Invert sugar is an equimolar mixture of glucose and fructose obtained by the hydrolysis of sucrose. It is called “invert” because sucrose is dextrorotatory (+66.5°), but the resulting mixture is levorotatory (-19.8°) because the levorotation of fructose is greater than the dextrorotation of glucose.
Why other options are incorrect:
Why other options are incorrect:
- • Lactose: Hydrolysis of lactose yields glucose and galactose, not invert sugar.
- • Sucrose: This is the starting material (table sugar). It only becomes “invert sugar” after it is broken down into its components.
- • Fructose: While fructose is a component of invert sugar, the term specifically refers to the 1:1 mixture of glucose and fructose combined.
Sucrose consists of:
- a) Glucose + glucose
- b) Glucose + fructose
- c) Glucose + galactose
- d) Glucose + mannose
Explanation:
Sucrose (table sugar) is a disaccharide composed of one molecule of $\alpha$-D-glucose and one molecule of $\beta$-D-fructose. They are linked together by an $\alpha(1 \rightarrow 2)$ glycosidic bond.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose + glucose: This combination forms Maltose (found in germinating cereals).
- • Glucose + galactose: This combination forms Lactose (found in milk).
- • Glucose + mannose: These are monosaccharide isomers, but they do not combine to form the standard nutritional disaccharide sucrose.
The monosaccharide units are linked by 1 → 4 glycosidic linkage in
- a) Maltose
- b) Sucrose
- c) Cellulose
- d) Cellobiose
Explanation:
Maltose consists of two glucose units linked by an α(1 → 4) glycosidic bond.
Why other options are incorrect:
Why other options are incorrect:
- • Sucrose: Glucose and fructose are linked by an α(1 → 2) glycosidic linkage.
- • Cellulose: Glucose units are linked by β(1 → 4) glycosidic bonds, but the question’s expected answer is Maltose.
- • Cellobiose: It also contains two glucose units linked by a β(1 → 4) glycosidic bond.
Which of the following is a non-reducing sugar?
- a) Isomaltose
- b) Maltose
- c) Lactose
- d) Trehalose
Explanation:
Trehalose is a non-reducing sugar because the glycosidic bond is formed between the anomeric carbons of the two glucose units ($\alpha1 \rightarrow 1$ linkage). Since both anomeric carbons are tied up in the bond, there is no free aldehyde or ketone group to act as a reducing agent.
Why other options are incorrect:
Why other options are incorrect:
- • Isomaltose: Has a free anomeric carbon at one end ($\alpha1 \rightarrow 6$ linkage), making it a reducing sugar.
- • Maltose: Contains one free anomeric carbon ($\alpha1 \rightarrow 4$ linkage), so it can reduce Fehling’s or Benedict’s solution.
- • Lactose: Milk sugar has a free anomeric carbon on the glucose residue ($\beta1 \rightarrow 4$ linkage), making it a reducing sugar.
Which of the following is a reducing sugar?
- a) Sucrose
- b) Trehalose
- c) Isomaltose
- d) Agar
Explanation:
Isomaltose is a reducing sugar. It consists of two glucose units linked by an $\alpha(1 \rightarrow 6)$ glycosidic bond. Because the second glucose unit has a free anomeric carbon (not involved in a bond), it can open into a straight chain to reveal a free aldehyde group, which can reduce alkaline copper or silver salts.
Why other options are incorrect:
Why other options are incorrect:
- • Sucrose: A non-reducing sugar because the linkage is between the anomeric carbons of both glucose (C1) and fructose (C2).
- • Trehalose: A non-reducing sugar because the linkage is between the C1 carbons of both glucose units, leaving no free anomeric centers.
- • Agar: This is a complex polysaccharide (a polymer) rather than a simple sugar. While it may have very few reducing ends, it is not classified as a “reducing sugar.”
A disaccharide formed by 1,1-glycosidic linkage between their monosaccharide units is:
- a) Lactose
- b) Maltose
- c) Trehalose
- d) Sucrose
Explanation:
Trehalose is unique among common disaccharides because it is formed by an $\alpha, \alpha$-1,1-glycosidic bond between two $\alpha$-glucose molecules. This means the linkage occurs between the anomeric carbons (C1) of both sugar units, making it a non-reducing sugar.
Why other options are incorrect :
Why other options are incorrect :
- • Lactose: Linked by a $\beta(1 \rightarrow 4)$ glycosidic bond between galactose and glucose.
- • Maltose: Linked by an $\alpha(1 \rightarrow 4)$ glycosidic bond between two glucose units.
- • Sucrose: Linked by an $\alpha(1 \rightarrow 2)$ glycosidic bond between glucose (C1) and fructose (C2).
Mutarotation refers to change in:
- a) pH
- b) Optical rotation
- c) Conductance
- d) Chemical properties
Explanation:
Mutarotation is the spontaneous change in the specific optical rotation of a solution of an optically active compound (like α-D-glucose or β-D-glucose) until it reaches a stable equilibrium value. This occurs due to the interconversion between the different anomeric forms of the sugar.
Why other options are incorrect:
Why other options are incorrect:
- • pH: Mutarotation is a structural change in the sugar molecule and does not involve the release or absorption of protons that would significantly change the pH.
- • Conductance: Sugars are non-electrolytes and do not dissociate into ions in water; therefore, their conductance does not change during this process.
- • Chemical properties: While the physical property (rotation) changes, the basic chemical properties (like being a reducing sugar) remain the same because the functional groups are largely unchanged.
A polysaccharide which is often called “animal starch” is:
- a) Glycogen
- b) Starch
- c) Inulin
- d) Dextrin
Explanation:
Glycogen is called “animal starch” because it serves as the primary storage form of energy (glucose) in animals and humans, much like starch does in plants. It is stored mainly in the liver and skeletal muscles. Structurally, it is highly branched, similar to amylopectin but with more frequent branching points.
Why other options are incorrect:
Why other options are incorrect:
- • Starch: This is the storage polysaccharide of plants, consisting of amylose and amylopectin.
- • Inulin: This is a polymer of fructose (fructosan) found in the bulbs and tubers of plants like dahlias and chicory; it is used to test kidney function (GFR).
- • Dextrin: These are a group of low-molecular-weight carbohydrates produced by the partial hydrolysis of starch.
A carbohydrate, commonly known as dextrose is:
- a) Dextrin
- b) D-Fructose
- c) D-Glucose
- d) Glycogen
Explanation:
D-Glucose is commonly called dextrose because it is dextrorotatory, meaning it rotates plane-polarized light to the right (+). It is the most common monosaccharide and the primary fuel for the body’s cells.
Why other options are incorrect:
Why other options are incorrect:
- • Dextrin: These are a group of low-molecular-weight carbohydrates produced by the hydrolysis of starch, not a single simple sugar.
- • D-Fructose: Commonly known as levulose because it is strongly levorotatory (rotates light to the left).
- • Glycogen: Known as “animal starch,” it is a large storage polysaccharide made of many glucose units.
The polysaccharide used in assessing the glomerular filtration rate (GFR) is:
- a) Glycogen
- b) Agar
- c) Inulin
- d) Hyaluronic acid
Explanation:
Inulin is a plant-derived polysaccharide (a polymer of fructose). It is used as the gold standard for measuring Glomerular Filtration Rate (GFR) because it is freely filtered by the glomeruli and is neither reabsorbed nor secreted by the renal tubules.
Why other options are incorrect:
Why other options are incorrect:
- • Glycogen: The storage form of glucose in animals; it is not used for renal function testing.
- • Agar: A complex polysaccharide from seaweed used primarily as a solidifying agent in laboratory culture media.
- • Hyaluronic acid: A glycosaminoglycan found in connective, epithelial, and neural tissues, acting as a lubricant in joints.
The constituent unit of inulin is:
- a) Glucose
- b) Fructose
- c) Mannose
- d) Galactose
Explanation:
Inulin is a fructosan, meaning it is a homopolysaccharide entirely composed of D-fructose units. These units are joined by $\beta(2 \rightarrow 1)$ glycosidic linkages. It is commonly found in the roots and tubers of plants like chicory and dahlias.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: This is the building block of starch, glycogen, and cellulose, not inulin.
- • Mannose: A sugar found in various plant gums and mucilages, but it does not form inulin.
- • Galactose: A component of milk sugar (lactose) and certain hemicelluloses, but not the constituent of inulin.
The polysaccharide found in the exoskeleton of invertebrates is:
- a) Pectin
- b) Chitin
- c) Cellulose
- d) Chondroitin sulphate
Explanation:
Chitin is a structural heteropolysaccharide composed of N-acetyl-D-glucosamine units. It is the primary component of the tough exoskeletons of arthropods (such as insects, crabs, and lobsters) and is also found in the cell walls of fungi.
Why other options are incorrect:
Why other options are incorrect:
- • Pectin: A structural heteropolysaccharide found in the primary cell walls and middle lamella of terrestrial plants, particularly fruits.
- • Cellulose: The most abundant organic polymer on Earth; it serves as the structural component of the cell wall of green plants and algae.
- • Chondroitin sulphate: A glycosaminoglycan found in vertebrate connective tissues, particularly in the cartilage, bone, and heart valves.
Which of the following is a heteroglycan?
- a) Dextrins
- b) Agar
- c) Inulin
- d) Chitin
Explanation:
A heteroglycan (heteropolysaccharide) is a polysaccharide composed of more than one type of monosaccharide unit. Agar is a heteroglycan because it is composed of D-galactose and 3,6-anhydro-L-galactose units.
Why other options are incorrect:
Why other options are incorrect:
- • Dextrins: These are homoglycans composed only of glucose units, formed during the hydrolysis of starch.
- • Inulin: This is a homoglycan composed entirely of fructose units (fructosan).
- • Chitin: This is a homoglycan composed of repeating units of N-acetyl-D-glucosamine.
The glycosaminoglycan which does not contain uronic acid is:
- a) Dermatan sulphate
- b) Chondroitin sulphate
- c) Keratan sulphate
- d) Heparan sulphate
Explanation:
Most glycosaminoglycans (GAGs) are composed of repeating disaccharide units consisting of an amino sugar and a uronic acid. Keratan sulphate is the exception; it contains D-galactose instead of a uronic acid in its repeating unit.
Why other options are incorrect:
Why other options are incorrect:
- • Dermatan sulphate: Contains L-iduronic acid (and occasionally D-glucuronic acid).
- • Chondroitin sulphate: Contains D-glucuronic acid as its uronic acid component.
- • Heparan sulphate: Contains both D-glucuronic acid and L-iduronic acid.
Keratan sulphate is found in abundance in:
- a) Heart muscle
- b) Liver
- c) Adrenal cortex
- d) Cornea
Explanation:
Keratan sulphate (specifically Keratan sulphate I) is found in high concentrations in the cornea of the eye, where it plays a crucial role in maintaining corneal transparency. It is also found in cartilage and bone (Keratan sulphate II).
Why other options are incorrect:
Why other options are incorrect:
- • Heart muscle: While the heart contains various glycosaminoglycans in its valves and connective tissue, Keratan sulphate is not the dominant feature of the muscle itself.
- • Liver: The liver is more associated with Heparan sulphate, which is found on the surfaces of hepatocytes.
- • Adrenal cortex: This is glandular tissue focused on steroid hormone production; it does not contain significant amounts of Keratan sulphate.
Repeating units of hyaluronic acid are:
- a) N-acetyl glucosamine and D-glucuronic acid
- b) N-acetyl galactosamine and D-glucuronic acid
- c) N-acetyl glucosamine and galactose
- d) N-acetyl galactosamine and L-iduronic acid
Explanation:
Hyaluronic acid is a major glycosaminoglycan found in the vitreous humor of the eye, synovial fluid, and umbilical cord. Its repeating disaccharide unit consists of D-glucuronic acid and N-acetyl glucosamine linked by $\beta(1 \rightarrow 3)$ and $\beta(1 \rightarrow 4)$ glycosidic bonds.
Why other options are incorrect:
Why other options are incorrect:
- • Option b: This combination (Galactosamine + Glucuronic acid) is found in Chondroitin sulphate.
- • Option c: This combination is found in Keratan sulphate (which lacks uronic acid).
- • Option d: This combination (Galactosamine + Iduronic acid) is characteristic of Dermatan sulphate.
The approximate number of branches in amylopectin is:
- a) 10
- b) 20
- c) 40
- d) 80
Explanation:
Amylopectin is a highly branched polysaccharide. Branching occurs via α(1→6) glycosidic bonds approximately every 20 to 30 glucose units along the main chain.
Why other options are incorrect:
Why other options are incorrect:
- • 10: This density of branching is more characteristic of Glycogen (which branches every 8-12 units) rather than amylopectin.
- • 40: This is too infrequent; amylopectin is more densely branched than once every 40 units.
- • 80: This value is much too high; molecules with such rare branching would behave more like the linear amylose.
In amylopectin, the intervals of glucose units of each branch is:
- a) 10–20
- b) 24–30
- c) 30–40
- d) 40–50
Explanation:
Amylopectin is a branched polysaccharide found in plants. While the main chain consists of α(1→4) glycosidic bonds, the branches occur via α(1→6) glycosidic bonds at intervals of approximately 24 to 30 glucose units.
Why other options are incorrect:
Why other options are incorrect:
- • 10–20: This range is closer to the branching frequency of Glycogen, which is more highly branched (every 8–12 units) than amylopectin.
- • 30–40 & 40–50: These intervals are too long. Amylopectin has a more compact structure than these ranges would suggest, requiring more frequent branching.
A polymer of glucose synthesized by the action of Leuconostoc mesenteroides in a sucrose medium is:
- a) Dextrans
- b) Dextrin
- c) Limit dextrin
- d) Inulin
Explanation:
Dextrans are complex, branched glucans (polymers of glucose) synthesized from sucrose by certain lactic acid bacteria, most notably Leuconostoc mesenteroides, using the enzyme dextransucrase.
Why other options are incorrect:
Why other options are incorrect:
- • Dextrin: These are low-molecular-weight carbohydrates produced by the hydrolysis of starch, not bacterial synthesis from sucrose.
- • Limit dextrin: This is the remaining core of a starch or glycogen molecule after it has been partially digested by amylase up to its branch points.
- • Inulin: This is a polysaccharide composed of fructose units (a fructan), typically found in the roots or rhizomes of plants like chicory and dahlias.
Glucose on reduction with sodium amalgam forms:
- a) Dulcitol
- b) Sorbitol
- c) Mannitol
- d) Mannitol and sorbitol
Explanation:
When Glucose (an aldose) is reduced by sodium amalgam (Na-Hg) and water, the aldehyde group at Carbon-1 is converted into a primary alcohol group. This process yields the sugar alcohol known as Sorbitol (also called Glucitol).
Why other options are incorrect:
Why other options are incorrect:
- • Dulcitol: This is the sugar alcohol formed by the reduction of Galactose.
- • Mannitol: This is the sugar alcohol formed by the reduction of Mannose.
- • Mannitol and sorbitol: This mixture is formed when Fructose (a ketose) is reduced, because the reduction of the carbonyl group at C2 creates a new chiral center, resulting in two epimers.
Glucose on oxidation does not give:
- a) Glycoside
- b) Glucosaccharic acid
- c) Gluconic acid
- d) Glucuronic acid
Explanation:
Oxidation of glucose typically results in various “sugar acids.” A Glycoside is not an oxidation product; it is formed by a condensation reaction between the anomeric hydroxyl group of glucose and an alcohol or another sugar molecule.
Why other options are incorrect:
Why other options are incorrect:
- • Glucosaccharic acid: Also known as Glucaric acid, this is formed when both the C1 aldehyde and C6 primary alcohol groups are oxidized (e.g., by concentrated nitric acid).
- • Gluconic acid: This is formed when only the C1 aldehyde group is oxidized (e.g., by mild oxidizing agents like bromine water).
- • Glucuronic acid: This is formed when only the C6 primary alcohol group is oxidized, often occurring through biological enzymatic pathways.
Oxidation of galactose with conc. HNO3 yields:
- a) Mucic acid
- b) Glucuronic acid
- c) Saccharic acid
- d) Gluconic acid
Explanation:
Concentrated Nitric Acid (HNO3) is a strong oxidizing agent that oxidizes both the aldehyde group (C1) and the primary alcohol group (C6) of a monosaccharide into carboxylic acid groups. When Galactose is treated this way, it forms galactaric acid, commonly known as Mucic acid. Mucic acid is clinically significant because it is insoluble in water and forms crystals, used as a test for the presence of galactose.
Why other options are incorrect:
Why other options are incorrect:
- • Glucuronic acid: This is a uronic acid formed when only the C6 of glucose is oxidized.
- • Saccharic acid: (Also known as Glucaric acid) This is the dicarboxylic acid formed by the oxidation of Glucose with conc. HNO3.
- • Gluconic acid: This is an aldonic acid formed when only the C1 aldehyde of glucose is oxidized (usually by a milder agent like bromine water).
A positive Benedict’s test is not given by:
- a) Sucrose
- b) Lactose
- c) Maltose
- d) Glucose
Explanation:
Benedict’s test is used to detect reducing sugars. Sucrose is a non-reducing sugar because the glycosidic bond is formed between the anomeric carbons of both glucose and fructose, leaving no free aldehyde or ketone group available to reduce the copper reagent.
Why other options are incorrect:
Why other options are incorrect:
- • Lactose: A reducing disaccharide because it has a free anomeric carbon on the glucose residue.
- • Maltose: A reducing disaccharide consisting of two glucose units with one free anomeric carbon.
- • Glucose: A monosaccharide with a free aldehyde group, making it a strong reducing sugar.
Starch is a:
- a) Polysaccharide
- b) Monosaccharide
- c) Disaccharide
- d) None of these
Explanation:
Starch is a polysaccharide composed of a large number of glucose units joined by glycosidic bonds. it is the primary energy storage molecule in plants and consists of two types of molecules: linear amylose and branched amylopectin.
Why other options are incorrect:
Why other options are incorrect:
- • Monosaccharide: These are simple sugars like Glucose or Fructose that cannot be hydrolyzed further into smaller sugars.
- • Disaccharide: These consist of exactly two monosaccharide units, such as Sucrose, Lactose, or Maltose.
- • None of these: This is incorrect as Starch clearly fits the definition of a polysaccharide.
A positive Seliwanoff’s test is obtained with:
- a) Glucose
- b) Fructose
- c) Lactose
- d) Maltose
Explanation:
Seliwanoff’s test is a chemical test used to distinguish aldoses from ketoses. Ketoses (like Fructose) are dehydrated more rapidly than aldoses when heated with concentrated HCl and resorcinol, resulting in a characteristic cherry-red color.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: This is an aldose; it reacts very slowly and does not produce a significant color change within the standard time limit of the test.
- • Lactose: A disaccharide composed of glucose and galactose, both of which are aldoses.
- • Maltose: A disaccharide composed of two glucose units, which are aldoses and will not yield a rapid positive result.
Osazones are not formed with:
- a) Glucose
- b) Fructose
- c) Sucrose
- d) Lactose
Explanation:
Osazone formation requires a free (or potentially free) carbonyl group (aldehyde or ketone) at C1 and C2. Sucrose is a non-reducing sugar where the glycosidic bond involves the anomeric carbons of both glucose and fructose. Since these functional groups are locked in the bond, sucrose cannot react with phenylhydrazine to form an osazone.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose & Fructose: Both are reducing monosaccharides that react with phenylhydrazine to form the same needle-shaped crystals (Glucosazone).
- • Lactose: It is a reducing disaccharide with a free anomeric carbon on the glucose unit, allowing it to form “powder-puff” or “hedgehog” shaped crystals (Lactosazone).
The most abundant carbohydrate found in nature is:
- a) Starch
- b) Glycogen
- c) Cellulose
- d) Chitin
Explanation:
Cellulose is the most abundant carbohydrate (and organic compound) on Earth. It is the primary structural component of the cell walls of green plants and many forms of algae.
Why other options are incorrect:
Why other options are incorrect:
- • Starch: While very common as a storage carbohydrate in plants, its total biomass is much less than the structural cellulose.
- • Glycogen: This is the storage form of glucose in animals and fungi; it is present in much smaller quantities globally compared to plant carbohydrates.
- • Chitin: Found in the exoskeletons of arthropods and cell walls of fungi, it is the second most abundant polysaccharide, but still ranks behind cellulose.
A carbohydrate found only in milk is:
- a) Glucose
- b) Galactose
- c) Lactose
- d) Maltose
Explanation:
Lactose is a disaccharide composed of glucose and galactose. It is synthesized in the mammary glands and is commonly known as “milk sugar” because it is found almost exclusively in the milk of mammals.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose: This is a universal monosaccharide found in blood, fruits, and as a building block for most polysaccharides.
- • Galactose: While it is a component of lactose, it is also found in various other biological molecules like glycolipids and glycoproteins throughout the body and in some plants.
- • Maltose: Also known as “malt sugar,” it is found in germinating cereals and is produced during the breakdown of starch.
An early feature of renal disease is:
- a) Impairment of the capacity of the tubule to perform osmotic work
- b) Decrease in maximal tubular excretory capacity
- c) Decrease in filtration factor
- d) Decrease in renal plasma flow
Explanation:
One of the earliest clinical signs of renal damage is the loss of the ability to concentrate or dilute urine. This process requires the renal tubules to perform “osmotic work” (moving solutes against a concentration gradient). When this capacity is impaired, it often precedes significant changes in GFR or blood flow.
Why other options are incorrect:
Why other options are incorrect:
- • Decrease in maximal tubular excretory capacity (Tm): This generally occurs as the disease progresses and a significant number of nephrons are lost or severely damaged.
- • Decrease in filtration factor: The filtration fraction (GFR/RPF) can be variable and is often maintained until later stages of kidney disease.
- • Decrease in renal plasma flow: While this does happen in renal disease, it is often a consequence of progressive vascular or structural damage rather than the very first functional indicator.
The ADH (Antidiuretic Hormone) test is based on the measurement of:
- a) Specific gravity of urine
- b) Concentration of urea in urine
- c) Concentration of urea in blood
- d) Volume of urine in ml/minute
Explanation:
ADH (Vasopressin) regulates the reabsorption of water in the kidneys. The ADH test (often as part of a water deprivation test) evaluates the kidney’s ability to concentrate urine. This is measured by the Specific Gravity or osmolality of the urine; a high specific gravity indicates that the kidneys are successfully reabsorbing water in response to ADH.
Why other options are incorrect:
Why other options are incorrect:
- • Concentration of urea in urine: Urea concentration varies based on protein intake and metabolism; it is not the primary indicator of ADH activity.
- • Concentration of urea in blood: This is a measure of general renal function (BUN) and GFR, rather than the specific water-regulating function of ADH.
- • Volume of urine: While ADH affects urine volume, the diagnostic assessment of ADH’s effectiveness is based on how concentrated the urine becomes, not just the total flow rate.
The specific gravity of urine normally ranges from:
- a) 0.900–0.999
- b) 1.003–1.030
- c) 1.000–1.001
- d) 1.101–1.120
Explanation:
The specific gravity of urine is a measure of the concentration of dissolved solutes. In a healthy individual with normal fluid intake, the value typically ranges between 1.003 and 1.030. This value varies throughout the day based on hydration, diet, and physical activity.
Why other options are incorrect:
Why other options are incorrect:
- • 0.900–0.999: These values are lower than the density of pure water (1.000), which is physically impossible for urine because it contains dissolved waste products.
- • 1.000–1.001: 1.000 is the specific gravity of distilled water. A value this low only occurs in cases of extreme over-hydration or conditions like Diabetes Insipidus.
- • 1.101–1.120: These values are excessively high and are not seen under normal physiological conditions; they may only occur after the administration of radiopaque contrast media.
Specific gravity of urine increases in:
- a) Diabetes mellitus
- b) Chronic glomerulonephritis
- c) Compulsive polydypsia
- d) Hypercalcemia
Explanation:
In Diabetes Mellitus, the presence of glucose in the urine (glycosuria) increases the concentration of dissolved substances, thereby raising the specific gravity despite the increased volume of urine.
Why other options are incorrect:
Why other options are incorrect:
- • Chronic glomerulonephritis: Usually leads to a “fixed” low specific gravity (isosthenuria) because the kidneys lose their ability to concentrate urine.
- • Compulsive polydypsia: Excessive water intake results in highly diluted urine, which significantly decreases specific gravity.
- • Hypercalcemia: Can lead to nephrogenic diabetes insipidus, impairing the kidney’s ability to concentrate urine, thus lowering specific gravity.
Fixation of specific gravity of urine to 1.010 is found in:
- a) Diabetes insipidus
- b) Compulsive polydypsia
- c) Cystinosis
- d) Chronic glomerulonephritis
Explanation:
Fixation of urine specific gravity at 1.010 is known as Isosthenuria. This occurs when the kidneys lose their ability to concentrate or dilute urine, resulting in urine that has the same osmolality as plasma. This is a characteristic sign of advanced Chronic glomerulonephritis or end-stage renal disease.
Why other options are incorrect:
Why other options are incorrect:
- • Diabetes insipidus: Characterized by Hyposthenuria (very low specific gravity, typically < 1.005) because the kidneys cannot reabsorb water.
- • Compulsive polydypsia: Leads to the excretion of large volumes of highly dilute urine with a very low specific gravity.
- • Cystinosis: While it can lead to renal damage (Fanconi syndrome), it is specifically associated with the accumulation of cystine; the specific “fixation” at 1.010 is the classic hallmark of generalized chronic renal failure.
Maltose can be formed by hydrolysis of:
- a) Starch
- b) Dextrin
- Glycogen
- d) All of these
Explanation:
Maltose is a disaccharide produced during the breakdown (hydrolysis) of long-chain glucose polymers. Starch, glycogen, and their intermediate breakdown products (dextrins) all consist of glucose units linked primarily by α(1→4) glycosidic bonds, which yield maltose when acted upon by enzymes like amylase.
Breakdown of Components:
Breakdown of Components:
- • Starch: The primary storage polysaccharide in plants; its digestion by salivary and pancreatic amylase produces maltose.
- • Glycogen: The storage polysaccharide in animals; similar to starch, its hydrolysis also results in maltose units.
- • Dextrin: These are low-molecular-weight carbohydrates produced by the partial hydrolysis of starch or glycogen; they are further hydrolyzed into maltose.
Number of stereoisomers of glucose is:
- a) 4
- b) 8
- c) 16
- d) None of these
Explanation:
The number of stereoisomers is calculated using the formula 2n, where ‘n’ is the number of asymmetric (chiral) carbon atoms. Glucose has 4 asymmetric carbon atoms (C-2, C-3, C-4, and C-5). Therefore, 24 = 16 stereoisomers.
Why other options are incorrect:
Why other options are incorrect:
- • 4: This would be the number of stereoisomers for a molecule with only 2 chiral centers (22=4).
- • 8: This is the number of stereoisomers for aldopentoses (like ribose), which have 3 chiral centers (23=8).
- • None of these: Incorrect because 16 is the mathematically proven number of stereoisomers for an aldohexose like glucose.
Fructose is present in hydrolysate of:
- a) Sucrose
- b) Inulin
- c) Both of the above
- d) None of these
Explanation:
Fructose is obtained when certain complex carbohydrates are broken down by water (hydrolysis). It is a constituent of both sucrose and inulin.
Why both are correct:
Why both are correct:
- • Sucrose: This is a disaccharide (common table sugar) composed of one glucose molecule and one fructose molecule.
- • Inulin: This is a polysaccharide (fructan) found in the roots/tubers of plants like dahlia and chicory; it is composed almost entirely of fructose units.
- • Result: Hydrolyzing either substance will release free fructose molecules.
A carbohydrate found in DNA is:
- a) Ribose
- b) Deoxyribose
- c) Ribulose
- d) All of these
Explanation:
DNA stands for Deoxyribonucleic Acid. Its structural backbone consists of a phosphate group and a five-carbon sugar (pentose) called 2-deoxyribose. It differs from ribose by the absence of an oxygen atom at the C-2 position.
Why other options are incorrect:
Why other options are incorrect:
- • Ribose: This pentose sugar is found in RNA (Ribonucleic Acid), not DNA.
- • Ribulose: This is a ketopentose sugar that acts as an intermediate in the Pentose Phosphate Pathway and the Calvin cycle in plants, but it is not a component of nucleic acids.
- • All of these: Incorrect, as only deoxyribose is the specific carbohydrate component of the DNA molecule.
Ribulose is a:
- a) Ketotetrose
- b) Aldotetrose
- c) Ketopentose
- d) Aldopentose
Explanation:
Ribulose is a monosaccharide containing five carbon atoms (pentose) and a ketone functional group (keto). It is an intermediate in the Pentose Phosphate Pathway and is essential in photosynthesis as Ribulose-1,5-bisphosphate.
Why other options are incorrect:
Why other options are incorrect:
- • Ketotetrose: A 4-carbon sugar with a ketone group (e.g., Erythrulose).
- • Aldotetrose: A 4-carbon sugar with an aldehyde group (e.g., Erythrose).
- • Aldopentose: A 5-carbon sugar with an aldehyde group (e.g., Ribose or Xylose).
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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