Chapter 100 – MCQs for Lab (Biochemistry)

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Chapter 100 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.

4951 to 5000 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 100 (Biochemistry)

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Questions 4951 to 5000

The apolipoprotein which forms the integral component of VLDL is:
  • a) B-100
  • b) B-48
  • c) A
  • d) D
Explanation: Apolipoprotein B-100 is the primary structural protein found in Very Low-Density Lipoproteins (VLDL), Intermediate-Density Lipoproteins (IDL), and Low-Density Lipoproteins (LDL). It is synthesized in the liver.
Why other options are incorrect:
  • • B-48: This is the integral apolipoprotein for Chylomicrons and is synthesized in the intestine.
  • • A (Apo A-I): This is the major structural protein found in HDL (High-Density Lipoprotein).
  • • D (Apo D): This is a glycoprotein primarily associated with HDL and is involved in lipid transport, but it is not an integral component of VLDL.
Serum LDL has been found to be increased in:
  • a) Obstructive jaundice
  • b) Hepatic jaundice
  • c) Hemolytic jaundice
  • d) Malabsorption syndrome
Explanation: In Obstructive jaundice (cholestasis), the excretion of cholesterol into the bile is blocked. This leads to a reflux of biliary lipids into the blood and the formation of an abnormal lipoprotein called Lipoprotein-X (Lp-X), which causes a significant rise in measured LDL and total cholesterol levels.
Why other options are incorrect:
  • • Hepatic jaundice: Liver cell damage usually results in decreased synthesis of lipoproteins, leading to lower or normal lipid levels.
  • • Hemolytic jaundice: This condition involves the rapid breakdown of red blood cells and primarily affects bilirubin levels, having no direct impact on LDL levels.
  • • Malabsorption syndrome: This leads to decreased serum LDL because the body is unable to absorb fats and lipids from the diet properly.
A lipoprotein associated with high incidence of coronary atherosclerosis is:
  • a) LDL
  • b) VLDL
  • c) IDL
  • d) HDL
Explanation: LDL (Low-Density Lipoprotein), often called “bad cholesterol,” is the primary carrier of cholesterol to peripheral tissues. High levels of LDL lead to its deposition in the arterial walls, where it becomes oxidized, leading to plaque formation (atherosclerosis) and coronary heart disease.
Why other options are incorrect:
  • • VLDL: While it carries triglycerides and can contribute to risk, it is not as directly or strongly associated with atherosclerosis as its metabolic product, LDL.
  • • IDL: This is a transient intermediate between VLDL and LDL; while pro-atherogenic, it is usually present in much smaller quantities than LDL.
  • • HDL: Known as “good cholesterol,” HDL is cardioprotective. It helps remove cholesterol from the arteries and transports it back to the liver (reverse cholesterol transport).
The apolipoprotein which acts as a ligand for the LDL receptor is:
  • a) B-48
  • b) B-100
  • c) A
  • d) C
Explanation: Apolipoprotein B-100 is the primary protein component of LDL. It serves as the specific ligand that binds to the LDL receptor (ApoB/E receptor) on the surface of liver and peripheral cells, allowing the cell to internalize the LDL particle via endocytosis.
Why other options are incorrect:
  • • B-48: Found in chylomicrons, B-48 is a truncated version of B-100. It lacks the C-terminal domain required to bind to the LDL receptor.
  • • A (Apo A-I): Primarily found in HDL; its main role is the activation of the enzyme LCAT (Lecithin-cholesterol acyltransferase) rather than binding to LDL receptors.
  • • C (Apo C-II): Mainly acts as an activator for the enzyme Lipoprotein Lipase (LPL) to break down triglycerides in VLDL and chylomicrons.
A lipoprotein inversely related to the incidence of coronary atherosclerosis is:
  • a) VLDL
  • b) IDL
  • c) LDL
  • d) HDL
Explanation: HDL (High-Density Lipoprotein) is considered “good cholesterol” because it is inversely related to the risk of coronary heart disease. It facilitates “Reverse Cholesterol Transport,” removing excess cholesterol from the arterial walls and transporting it back to the liver for excretion.
Why other options are incorrect:
  • • VLDL: High levels are directly (not inversely) associated with an increased risk of cardiovascular disease.
  • • IDL: As a pro-atherogenic intermediate, its presence in high amounts is positively correlated with plaque formation.
  • • LDL: Known as “bad cholesterol,” its relationship to atherosclerosis is direct; high levels significantly increase the risk.
The primary biochemical lesion in homozygote with familial hypercholesterolemia (type IIa) is:
  • a) Loss of feedback inhibition of HMG reductase
  • b) Loss of apolipoprotein B
  • c) Increased production of LDL from VLDL
  • d) Functional deficiency of plasma membrane receptors for LDL
Explanation: Familial Hypercholesterolemia (Type IIa) is primarily caused by a genetic defect (mutation) in the gene encoding the LDL receptor. In homozygotes, these receptors are either completely absent or non-functional, preventing the liver and peripheral tissues from clearing LDL from the blood, leading to extremely high cholesterol levels from birth.
Why other options are incorrect:
  • • Loss of feedback inhibition: While HMG-CoA reductase remains active (because cholesterol isn’t entering the cell to inhibit it), this is a secondary consequence of the receptor defect, not the primary lesion.
  • • Loss of apolipoprotein B: In Type IIa, Apo B-100 levels are actually increased because LDL (which contains Apo B-100) is not being cleared.
  • • Increased production: The massive rise in LDL is due to a failure in clearance (removal) from the plasma, rather than an overproduction from VLDL.
In abetalipoproteinemia, the biochemical defect is in:
  • a) Apo-B synthesis
  • b) Lipoprotein lipase activity
  • c) Cholesterol ester hydrolase
  • d) LCAT activity
Explanation: Abetalipoproteinemia is a rare genetic disorder characterized by the inability to synthesize/assemble lipoproteins containing Apolipoprotein B (specifically Apo B-48 in the intestine and Apo B-100 in the liver). This results in the near-total absence of chylomicrons, VLDL, and LDL in the plasma.
Why other options are incorrect:
  • • Lipoprotein lipase activity: Deficiency of this enzyme leads to Type I Hyperlipoproteinemia (Hyperchylomicronemia), not the absence of lipoproteins.
  • • Cholesterol ester hydrolase: This enzyme is involved in the breakdown of cholesterol esters within cells or lysosomes; its deficiency is seen in Wolman disease.
  • • LCAT activity: Lecithin-cholesterol acyltransferase (LCAT) deficiency leads to Norum disease, which primarily affects HDL maturation and results in “fish-eye” corneal opacities.
Familial hypertriaacylglycerolemia (Type IV Hyperlipoproteinemia) is associated with:
  • a) Over production of VLDL
  • b) Increased LDL concentration
  • c) Increased HDL concentration
  • d) Slow clearance of chylomicrons
Explanation: Familial Hypertriaacylglycerolemia (Type IV) is a common condition characterized by elevated fasting plasma triglycerides. The primary defect is the hepatic overproduction of VLDL (Very Low-Density Lipoprotein) or a decreased clearance of VLDL from the blood.
Why other options are incorrect:
  • • Increased LDL: This is characteristic of Type IIa hyperlipoproteinemia (Familial Hypercholesterolemia).
  • • Increased HDL: This is generally a cardioprotective state and is not associated with hypertriaacylglycerolemia; in fact, Type IV patients often have low HDL.
  • • Slow clearance of chylomicrons: This is the hallmark of Type I hyperlipoproteinemia (Familial Chylomicronemia), usually caused by a deficiency in Lipoprotein Lipase (LPL).
For synthesis of prostaglandins, the essential fatty acids give rise to a fatty acid containing:
  • a) 12 carbon atoms
  • b) 16 carbon atoms
  • c) 20 carbon atoms
  • d) 24 carbon atoms
Explanation: Prostaglandins are members of a group of lipid compounds called eicosanoids. The term “eicosa” is Greek for twenty, referring to the fact that these molecules are synthesized from 20-carbon polyunsaturated fatty acids, most commonly Arachidonic Acid.
Why other options are incorrect:
  • • 12 carbon atoms: Fatty acids like Lauric acid have 12 carbons but are not precursors for signaling molecules like prostaglandins.
  • • 16 carbon atoms: Palmitic acid is a common 16-carbon saturated fatty acid, but it cannot be used to synthesize eicosanoids.
  • • 24 carbon atoms: Very long-chain fatty acids like Lignoceric acid (24 carbons) are found in brain fats (sphingolipids) but are not used for prostaglandin synthesis.
All active prostaglandins have at least one double bond between positions:
  • a) 7 and 8
  • b) 10 and 11
  • c) 13 and 14
  • d) 16 and 17
Explanation: Naturally occurring prostaglandins are grouped into three series (1, 2, and 3) based on the number of double bonds in their side chains. All physiologically active prostaglandins contain a trans-double bond at the C13-C14 position.
Why other options are incorrect:
  • • 7 and 8: Double bonds are not typically found at this position in the major active prostaglandin series.
  • • 10 and 11: These carbons are part of the cyclopentane ring structure; while substitutions (like hydroxyl or keto groups) occur here, a double bond is not a universal feature of all active PGs at this site.
  • • 16 and 17: A double bond at C17-C18 is found in the 3-series (derived from EPA), but it is not present in the 1-series or 2-series prostaglandins.
Normal range of plasma total phospholipids is:
  • a) 0.2–0.6 mmol/L
  • b) 0.9–2.0 mmol/L
  • c) 1.8–5.8 mmol/L
  • d) 2.8–5.3 mmol/L
Explanation: The normal range for total phospholipids in human plasma is approximately 1.8–5.8 mmol/L (which corresponds to about 150–450 mg/dL). Phospholipids, such as lecithin and sphingomyelin, are essential components of lipoproteins and help emulsify other lipids for transport.
Why other options are incorrect:
  • • 0.2–0.6 mmol/L: This range is more characteristic of fasting Free Fatty Acids (NEFA) levels in the plasma.
  • • 0.9–2.0 mmol/L: This represents a typical normal range for fasting Triglycerides.
  • • 2.8–5.3 mmol/L: This range is generally associated with the reference values for Total Cholesterol in healthy adults.
HDL2 have the density in the range of:
  • a) 1.006–1.019
  • b) 1.019–1.032
  • c) 1.032–1.063
  • d) 1.063–1.125
Explanation: HDL (High-Density Lipoprotein) is divided into two major sub-fractions: HDL2 and HDL3. HDL2 is larger, contains more lipids, and is less dense than HDL3. Its density range is 1.063–1.125 g/mL.
Why other options are incorrect:
  • • 1.006–1.019: This is the density range for IDL (Intermediate-Density Lipoprotein).
  • • 1.019–1.063: (Covering options b and c) This is the density range for LDL (Low-Density Lipoprotein).
  • • HDL3 (Note): For comparison, HDL3 is the densest fraction, ranging from 1.125–1.210 g/mL.
β-lipoproteins have the density in the range of:
  • a) 0.95–1.006
  • b) 1.006–1.019
  • c) 1.019–1.063
  • d) 1.063–1.125
Explanation: β-lipoproteins is the electrophoretic term for LDL (Low-Density Lipoprotein). In the ultracentrifugation method, LDL is defined by its density range of 1.019–1.063 g/mL.
Why other options are incorrect:
  • • 0.95–1.006: This is the density range for VLDL (also known as Pre-β lipoprotein).
  • • 1.006–1.019: This is the density range for IDL (Intermediate-Density Lipoprotein).
  • • 1.063–1.125: This is the density range for HDL2 (a sub-fraction of α-lipoprotein).
IDL (Intermediate-Density Lipoprotein) has a density in the range of:
  • a) 0.95–1.006
  • b) 1.006–1.019
  • c) 1.019–1.032
  • d) 1.032–1.163
Explanation: IDL (Intermediate-Density Lipoprotein) is formed during the degradation of VLDL. It serves as a transient state between VLDL and LDL. Based on ultracentrifugation, its density falls precisely between the two, in the range of 1.006–1.019 g/mL.
Why other options are incorrect:
  • • 0.95–1.006: This is the density range for VLDL (Very Low-Density Lipoprotein).
  • • 1.019–1.032: This is the density range for LDL1 (a lighter sub-fraction of LDL).
  • • 1.032–1.163: This range overlaps with LDL2 (heavy LDL) and HDL2, representing much denser particles.
Aspirin inhibits the activity of the enzyme:
  • a) Lipoxygenase
  • b) Cyclooxygenase
  • c) Phospholipase A1
  • d) Phospholipase A2
Explanation: Aspirin (acetylsalicylic acid) acts by irreversibly acetylating a serine residue in the active site of Cyclooxygenase (COX-1 and COX-2). This inhibition prevents the conversion of arachidonic acid into prostaglandins and thromboxanes, which are mediators of inflammation and platelet aggregation.
Why other options are incorrect:
  • • Lipoxygenase: This enzyme is responsible for the synthesis of leukotrienes. Aspirin does not inhibit it; in some cases, aspirin can cause arachidonic acid to be shunted toward this pathway.
  • • Phospholipase A1: This enzyme cleaves the acyl chain from the first carbon of a phospholipid and is not a target for aspirin.
  • • Phospholipase A2: This enzyme releases arachidonic acid from membrane phospholipids. It is inhibited by corticosteroids (via lipocortins), not by aspirin.
A “suicide enzyme” is:
  • a) Cycloxygenase
  • b) Lipooxygenase
  • c) Phospholipase A1
  • d) Phospholipase A2
Explanation: Cycloxygenase (COX), also known as Prostaglandin H synthase, is frequently referred to as a “suicide enzyme.” This is because the enzyme possesses the unique property of self-catalyzed inactivation; it undergoes a process where the catalytic activity is destroyed by the very products or free radicals generated during its own reaction.
Why other options are incorrect:
  • • Lipooxygenase: While involved in the eicosanoid pathway (producing leukotrienes), it is not typically characterized by the same rapid self-inactivation mechanism as COX.
  • • Phospholipase A1 & A2: These are hydrolases that release fatty acids from phospholipids. They follow standard catalytic cycles and do not “commit suicide” or self-destruct during their normal function.
In adipose tissue, prostaglandins decrease:
  • a) Lipogenesis
  • b) Lipolysis
  • c) Gluconeogenesis
  • d) Glycogenolysis
Explanation: Prostaglandins (specifically the PGE series) are known to inhibit the activity of the enzyme adenylate cyclase in adipose tissue. This leads to a decrease in cyclic AMP (cAMP) levels, which in turn prevents the activation of Hormone-Sensitive Lipase (HSL), thereby decreasing lipolysis (the breakdown of fats).
Why other options are incorrect:
  • • Lipogenesis: This is the synthesis of fatty acids; it is primarily stimulated by insulin, and prostaglandins do not have a primary inhibitory effect on this process.
  • • Gluconeogenesis: This is the synthesis of glucose from non-carbohydrate sources, a process that occurs mainly in the liver and kidneys, not in adipose tissue.
  • • Glycogenolysis: This is the breakdown of glycogen into glucose, which occurs primarily in the liver and skeletal muscle.
The optimal pH for the enzyme pepsin is:
  • a) 1.0–2.0
  • b) 4.0–5.0
  • c) 5.2–6.0
  • d) 5.8–6.0
Explanation: Pepsin is a gastric protease secreted by the chief cells of the stomach as the inactive zymogen, pepsinogen. It requires a highly acidic environment for both its activation and its catalytic activity, with an optimal pH range of 1.5 to 2.0.
Why other options are incorrect:
  • • 4.0–5.0: At this pH level, the activity of pepsin significantly declines as the enzyme begins to lose its optimal conformation.
  • • 5.2–6.0 & 5.8–6.0: Pepsin becomes increasingly inactive as the pH rises toward neutrality. Once the pH reaches 7.0 or higher (as it does in the small intestine), pepsin is irreversibly denatured and inactivated.
Pepsinogen is converted to active pepsin by:
  • a) HCl
  • b) Bile salts
  • c) Ca++
  • d) Enterokinase
Explanation: HCl (Hydrochloric acid), secreted by the parietal cells of the stomach, creates the low pH necessary to cleave a protective polypeptide from pepsinogen to convert it into active pepsin. Once some pepsin is formed, it can also activate more pepsinogen molecules, a process known as autocatalysis.
Why other options are incorrect:
  • • Bile salts: These are produced by the liver and involved in the emulsification of fats in the small intestine; they do not activate gastric enzymes.
  • • Ca++: While calcium ions are essential for some enzymes (like α-amylase or certain clotting factors), they are not the primary trigger for pepsinogen activation.
  • • Enterokinase: This enzyme is found in the duodenum and is responsible for activating trypsinogen into trypsin, not pepsinogen.
The optimal pH for the enzyme rennin is:
  • a) 2.0
  • b) 4.0
  • c) 8.0
  • d) 6.0
Explanation: Rennin (also known as chymosin) is a proteolytic enzyme found in the gastric juice of infants. It is responsible for the curdling of milk (converting casein to paracasein). Its optimal activity occurs at a pH of about 4.0, which matches the less acidic environment of an infant’s stomach compared to an adult’s.
Why other options are incorrect:
  • • 2.0: This is the optimal pH for pepsin, which is the primary protein-digesting enzyme in the adult stomach.
  • • 8.0: This is an alkaline pH, typical of the environment in the small intestine where enzymes like trypsin and chymotrypsin operate.
  • • 6.0: While closer to the optimal range than 8.0, rennin activity is significantly higher and more efficient at the slightly more acidic pH of 4.0.
The optimal pH for the enzyme trypsin is:
  • a) 1.0–2.0
  • b) 2.0–4.0
  • c) 5.2–6.2
  • d) 7.5–8.5
Explanation: Trypsin is a pancreatic serine protease that functions in the small intestine. Because the environment of the small intestine is neutralized by bicarbonate, trypsin requires an alkaline (basic) environment to function, with an optimal pH of approximately 7.8 to 8.5.
Why other options are incorrect:
  • • 1.0–2.0: This is the optimal pH for pepsin in the stomach; trypsin would be denatured at this high acidity.
  • • 2.0–4.0: This range is still too acidic for pancreatic enzymes, which are designed to work after gastric acid has been neutralized.
  • • 5.2–6.2: While closer to neutral, trypsin activity is significantly lower in the slightly acidic range and reaches its peak only when the environment is clearly alkaline.
Trypsinogen is converted to active trypsin by:
  • a) Enterokinase
  • b) Bile salts
  • c) HCl
  • d) Mg++
Explanation: Enterokinase (also known as enteropeptidase) is an enzyme produced by the cells of the duodenum. It initiates protein digestion by cleaving a hexapeptide from the N-terminal end of trypsinogen, converting it into active trypsin. Once trypsin is formed, it can also activate more trypsinogen molecules (autocatalysis).
Why other options are incorrect:
  • • Bile salts: These are detergents synthesized in the liver that help in the emulsification and absorption of dietary fats, but they do not activate proteolytic enzymes.
  • • HCl: Hydrochloric acid is responsible for activating pepsinogen in the stomach. Trypsinogen activation occurs in the alkaline environment of the small intestine.
  • • Mg++: Magnesium ions are cofactors for many enzymes (like kinases), but they are not the specific biochemical trigger for the conversion of trypsinogen to trypsin.
Pepsin acts on denatured proteins to produce:
  • a) Proteoses and peptones
  • b) Polypeptides
  • c) Peptides
  • d) Dipeptide
Explanation: Pepsin is an endopeptidase that begins the process of protein digestion in the stomach. It hydrolyzes internal peptide bonds of large protein molecules, breaking them down into smaller, but still relatively large, fragments known as proteoses and peptones.
Why other options are incorrect:
  • • Polypeptides/Peptides: While these terms are general, “proteoses and peptones” are the specific classical terms used to describe the initial intermediate products of gastric protein digestion.
  • • Dipeptide: Proteins are not broken down into dipeptides in the stomach. This level of digestion occurs later in the small intestine through the action of pancreatic enzymes and intestinal brush border peptidases.
Rennin converts casein to paracasein in the presence of:
  • a) Ca++
  • b) Mg++
  • c) Na+
  • d) K+
Explanation: Rennin (chymosin) acts on the milk protein casein to convert it into paracasein. In the presence of Calcium ions (Ca++), paracasein is further converted into an insoluble precipitate called calcium paracaseinate, which results in the “curdling” or clotting of milk.
Why other options are incorrect:
  • • Mg++: Magnesium ions are not the physiological requirement for the formation of the insoluble milk curd during rennin digestion.
  • • Na+: Sodium is a monovalent cation and does not facilitate the bridging required to form the insoluble paracaseinate complex.
  • • K+: Potassium, like sodium, does not play a role in the structural formation of the milk clot.
An exopeptidase is:
  • a) Trypsin
  • b) Chymotrypsin
  • c) Elastase
  • d) Carboxypeptidase
Explanation: Exopeptidases are enzymes that catalyze the cleavage of the terminal peptide bond (at the ends of the chain). Carboxypeptidase is a classic example of an exopeptidase as it removes amino acids one by one from the C-terminal end of a polypeptide.
Why other options are incorrect:
  • • Trypsin: This is an endopeptidase, meaning it breaks internal peptide bonds (specifically after Lysine or Arginine).
  • • Chymotrypsin: This is also an endopeptidase that acts on internal bonds (specifically after aromatic amino acids).
  • • Elastase: Like the others above, this is an endopeptidase that acts on internal bonds of proteins like elastin.
The enzyme trypsin is specific for peptide bonds of:
  • a) Basic amino acids
  • b) Acidic amino acids
  • c) Aromatic amino acids
  • d) Next to small amino acid residues
Explanation: Trypsin is an endopeptidase that specifically hydrolyzes peptide bonds where the carboxyl group is contributed by the basic amino acids, namely Lysine and Arginine. This specificity is due to a negatively charged aspartate residue at the bottom of the enzyme’s binding pocket which attracts the positively charged side chains of these amino acids.
Why other options are incorrect:
  • • Acidic amino acids: Amino acids like Aspartate and Glutamate are not targets for trypsin; enzymes like Pepsin or specific endopeptidases may act near them.
  • • Aromatic amino acids: Peptide bonds involving Phenylalanine, Tyrosine, and Tryptophan are the primary targets of Chymotrypsin.
  • • Next to small amino acid residues: This describes the specificity of Elastase, which cleaves bonds next to small, uncharged amino acids like Alanine, Glycine, or Serine.
Chymotrypsin is specific for peptide bonds containing:
  • a) Uncharged amino acid residues
  • b) Acidic amino acids
  • c) Basic amino acids
  • d) Small amino acid residues
Explanation: Chymotrypsin is a pancreatic endopeptidase that specifically hydrolyzes peptide bonds where the carboxyl group is contributed by large uncharged (specifically hydrophobic and aromatic) amino acids such as Phenylalanine, Tyrosine, and Tryptophan. Its active site contains a hydrophobic pocket that accommodates these bulky rings.
Why other options are incorrect:
  • • Acidic amino acids: Bonds involving Aspartate or Glutamate are not the primary target for chymotrypsin.
  • • Basic amino acids: These (Lysine and Arginine) are the specific targets for Trypsin.
  • • Small amino acid residues: Peptide bonds involving small amino acids like Alanine, Glycine, or Serine are the specific targets for Elastase.
The end product of protein digestion in the G.I.T. is:
  • a) Dipeptide
  • b) Tripeptide
  • c) Polypeptide
  • d) Amino acid
Explanation: The ultimate goal of protein digestion in the Gastrointestinal Tract (G.I.T.) is to break down complex proteins into their simplest building blocks: amino acids. These are the primary forms in which the body can absorb nutrients into the portal bloodstream through the intestinal mucosa.
Why other options are incorrect:
  • • Polypeptide: These are large chains of amino acids and represent an early stage of digestion (mostly occurring in the stomach).
  • • Dipeptide & Tripeptide: While some di- and tripeptides are absorbed into the intestinal cells (enterocytes), they are almost entirely hydrolyzed into free amino acids by intracellular peptidases before entering the blood.
Natural L-isomers of amino acids are absorbed from the intestine by:
  • a) Passive diffusion
  • b) Simple diffusion
  • c) Facilitated diffusion
  • d) Active process
Explanation: The absorption of natural L-amino acids occurs primarily via an active process. This is a carrier-mediated, secondary active transport mechanism that is dependent on a sodium (Na+) gradient.
Why other options are incorrect:
  • • Passive/Simple diffusion: These processes do not require energy or carriers. While some D-amino acids might be absorbed this way, L-amino acids require active transport to move against a concentration gradient.
  • • Facilitated diffusion: Although carrier proteins are involved, facilitated diffusion does not require energy (ATP). The absorption of L-amino acids is specifically an energy-dependent active process.
Abnormalities of blood clotting include:
  • a) Haemophilia
  • b) Christmas disease
  • c) Gout
  • d) Both (Haemophilia) and (Christmas disease)
Explanation: Both Haemophilia (Haemophilia A) and Christmas disease (Haemophilia B) are genetic disorders that impair the blood’s ability to clot. Haemophilia A is a deficiency of Factor VIII, while Christmas disease is a deficiency of Factor IX.
Why other options are incorrect:
  • • Gout: This is a metabolic disorder characterized by the accumulation of uric acid crystals in the joints, leading to inflammation and arthritis; it is not a clotting disorder.
  • • Both (Haemophilia) and (Christmas disease): This is the correct choice because both “a” and “b” represent conditions where blood fails to clot normally.
An important reaction for the synthesis of amino acid from carbohydrate intermediates is transamination which requires the cofactor:
  • a) Thiamin
  • b) Riboflavin
  • c) Niacin
  • d) Pyridoxal phosphate
Explanation: Pyridoxal phosphate (PLP), which is the active form of Vitamin B6, is the essential cofactor for all transamination reactions. It acts as a temporary carrier of the amino group during its transfer from an amino acid to a keto acid.
Why other options are incorrect:
  • • Thiamin: As TPP, it is mainly involved in oxidative decarboxylation (like the pyruvate dehydrogenase complex) and the pentose phosphate pathway, not transamination.
  • • Riboflavin: As FAD/FMN, it is primarily involved in oxidation-reduction (redox) reactions in the electron transport chain.
  • • Niacin: As NAD/NADP, it is also involved in redox reactions and acts as an electron carrier, but does not participate in the transfer of amino groups.
The main sites for oxidative deamination are:
  • a) Liver and kidney
  • b) Skin and pancreas
  • c) Intestine and mammary gland
  • d) Lung and spleen
Explanation: Oxidative deamination is the process where amino acids are broken down by removing the amino group to form keto acids and ammonia. This occurs primarily in the liver and kidney, specifically within the mitochondria, using the enzyme glutamate dehydrogenase.
Why other options are incorrect:
  • • Skin and pancreas: The skin serves as a protective barrier and the pancreas is involved in digestion and glucose regulation; neither is a primary site for amino acid deamination.
  • • Intestine and mammary gland: The intestine is focused on nutrient absorption and mammary glands on milk production, not the systemic removal of amino groups.
  • • Lung and spleen: The lungs are for gas exchange and the spleen for blood filtration and immune response; they do not perform oxidative deamination.
A positive nitrogen balance occurs:
  • a) In growing infant
  • b) Following surgery
  • c) In advanced cancer
  • d) In kwashiorkor
Explanation: A positive nitrogen balance occurs when the intake of nitrogen (protein) into the body is greater than the amount excreted. This happens during periods of rapid growth, such as in growing infants, children, and during pregnancy, as the body is actively building new tissues.
Why other options are incorrect:
  • • Following surgery: The body is in a catabolic (breakdown) state due to stress and tissue repair needs, often leading to a negative nitrogen balance.
  • • In advanced cancer: Cancer causes metabolic wasting (cachexia) and tissue breakdown, resulting in a negative nitrogen balance.
  • • In kwashiorkor: This is a state of severe protein malnutrition where nitrogen intake is extremely low, leading to a negative nitrogen balance.
The main site of urea synthesis in mammals is:
  • a) Liver
  • b) Skin
  • c) Intestine
  • d) Kidney
Explanation: The Liver is the primary site of urea synthesis. The urea cycle (also known as the Ornithine cycle) takes place almost exclusively in hepatocytes (liver cells) to convert highly toxic ammonia into less toxic urea for excretion.
Why other options are incorrect:
  • • Skin: While the skin can excrete small amounts of urea through sweat, it does not possess the enzymes required to synthesize urea.
  • • Intestine: The intestine is a site where ammonia is produced by bacterial action on proteins, but this ammonia must travel to the liver to be converted into urea.
  • • Kidney: The kidneys are the main site for urea excretion (filtering it from the blood into urine), but they do not synthesize the majority of the body’s urea.
The enzymes of urea synthesis are found in:
  • a) Mitochondria only
  • b) Cytosol only
  • c) Both mitochondria and cytosol
  • d) Nucleus
Explanation: The urea cycle is a compartmentalized metabolic pathway. The first two reactions occur in the mitochondria (formation of carbamoyl phosphate and citrulline), while the remaining reactions occur in the cytosol.
Why other options are incorrect:
  • • Mitochondria only: This is incorrect because the enzymes Argininosuccinate synthetase, Argininosuccinate lyase, and Arginase are located in the cytosol.
  • • Cytosol only: This is incorrect because the cycle must begin in the mitochondria with the enzyme Carbamoyl phosphate synthetase I.
  • • Nucleus: The nucleus houses genetic material and is not a site for the metabolic reactions of the urea cycle.
The number of ATP molecules required for the synthesis of one molecule of urea is:
  • a) 0
  • b) 1
  • c) 2
  • d) 3
Explanation: The synthesis of one molecule of urea requires 3 ATP molecules. However, it consumes 4 high-energy phosphate bonds because one ATP is hydrolyzed to AMP and inorganic pyrophosphate (PPi).

• 2 ATP are used to form Carbamoyl Phosphate.
• 1 ATP is used to form Argininosuccinate.
Why other options are incorrect:
  • • 0, 1, or 2: These values are insufficient. The urea cycle is an energy-intensive process that requires 3 ATP molecules to process toxic ammonia into urea.
Most of the ammonia released from L-α-amino acids reflects the coupled action of transaminase and:
  • a) L-glutamate dehydrogenase
  • b) L-amino acid oxidase
  • c) Histidase
  • d) Serine dehydratase
Explanation: This coupled process is known as transdeamination. Most amino acids transfer their amino group to α-ketoglutarate via transamination to form glutamate. Then, L-glutamate dehydrogenase performs oxidative deamination on glutamate to release the nitrogen as free ammonia.
Why other options are incorrect:
  • • L-amino acid oxidase: While this enzyme can release ammonia directly from amino acids, its activity is very low in tissues and accounts for very little total ammonia production.
  • • Histidase: This enzyme is specific only to the metabolism of the amino acid Histidine, not the general pool of L-α-amino acids.
  • • Serine dehydratase: This enzyme specifically catalyzes the deamination of Serine (and Threonine), but it is not the primary mechanism for ammonia release from the majority of other amino acids.
In urea synthesis, the amino acid derivative functioning solely as an essential enzyme activator is:
  • a) N-acetyl glutamate
  • b) Ornithine
  • c) Citrulline
  • d) Arginine
Explanation: N-acetyl glutamate (NAG) is the essential allosteric activator for the enzyme Carbamoyl Phosphate Synthetase I (CPS-I), which is the rate-limiting step of the urea cycle. Without NAG, CPS-I remains inactive, and urea synthesis cannot proceed.
Why other options are incorrect:
  • • Ornithine: It acts as a substrate/carrier in the cycle that combines with carbamoyl phosphate to form citrulline.
  • • Citrulline: It is an intermediate product formed in the mitochondria and transported to the cytosol.
  • • Arginine: While Arginine is an intermediate of the cycle, it also stimulates the production of NAG, but it is not the direct activator of the CPS-I enzyme itself.
The enzyme carbamoyl phosphate synthetase requires:
  • a) Mg++
  • b) Ca++
  • c) Na+
  • d) K+
Explanation: Magnesium ions (Mg++) are essential cofactors for the enzyme carbamoyl phosphate synthetase I. Since this enzyme requires the hydrolysis of ATP to provide energy for the reaction, Mg++ is necessary to complex with ATP and facilitate the transfer of phosphate groups.
Why other options are incorrect:
  • • Ca++: Calcium often acts as a signaling molecule or a cofactor for different enzymes (like those in blood clotting), but it is not the primary ion for ATP-dependent urea cycle enzymes.
  • • Na+ and K+: These are monovalent cations primarily involved in maintaining osmotic balance and membrane potential, rather than acting as enzymatic cofactors for the urea cycle.
Control of the urea cycle involves the enzyme:
  • a) Carbamoyl phosphate synthetase
  • b) Ornithine transcarbamoylase
  • c) Argininosuccinase
  • d) Arginase
Explanation: Carbamoyl phosphate synthetase I (CPS-I) is the rate-limiting and primary regulatory enzyme of the urea cycle. It is allosterically controlled by N-acetylglutamate (NAG); the cycle cannot proceed efficiently without the activation of this specific enzyme.
Why other options are incorrect:
  • • Ornithine transcarbamoylase: While it is the second enzyme in the cycle, it is not the primary site of metabolic control.
  • • Argininosuccinase: This enzyme catalyzes the cleavage of argininosuccinate into arginine and fumarate in the cytosol; it is an essential step but not a regulatory one.
  • • Arginase: This is the final enzyme of the cycle that releases urea and regenerates ornithine. Its activity is essential for completing the cycle but it does not serve as the main control point.
Transfer of the carbamoyl moiety of carbamoyl phosphate to ornithine is catalysed by a liver mitochondrial enzyme:
  • a) Carbamoyl phosphate synthetase
  • b) Ornithine transcarbamoylase
  • c) N-acetyl glutamate synthetase
  • d) N-acetyl glutamate hydrolase
Explanation: Ornithine transcarbamoylase (OTC) is the mitochondrial enzyme that catalyzes the transfer of the carbamoyl group from carbamoyl phosphate to the amino acid ornithine, resulting in the formation of citrulline.
Why other options are incorrect:
  • • Carbamoyl phosphate synthetase: This enzyme produces carbamoyl phosphate from ammonia and bicarbonate; it does not transfer the moiety to ornithine.
  • • N-acetyl glutamate synthetase: This enzyme produces N-acetylglutamate (the activator for CPS-I) from acetyl-CoA and glutamate.
  • • N-acetyl glutamate hydrolase: This enzyme is involved in the breakdown of N-acetylglutamate, not the urea cycle steps themselves.
A compound serving as a link between the citric acid cycle and the urea cycle is:
  • a) Malate
  • b) Citrate
  • c) Succinate
  • d) Fumarate
Explanation: Fumarate is the molecule that provides a direct link between the two cycles (often called the “Krebs Bicycle”). In the urea cycle, the enzyme argininosuccinate lyase cleaves argininosuccinate into arginine and fumarate. The fumarate can then enter the Citric Acid Cycle (TCA cycle).
Why other options are incorrect:
  • • Malate: While fumarate is converted to malate within the TCA cycle, it is fumarate itself that is the direct byproduct of the urea cycle reaction.
  • • Citrate: This is the first intermediate of the citric acid cycle but does not play a direct role in the reactions of the urea cycle.
  • • Succinate: This is a TCA cycle intermediate that occurs after alpha-ketoglutarate, but it is not produced during the steps of urea synthesis.
The 2 nitrogen atoms in urea are contributed by:
  • a) Ammonia and glutamate
  • b) Glutamine and glutamate
  • c) Ammonia and aspartate
  • d) Ammonia and alanine
Explanation: The urea molecule ($NH_2-CO-NH_2$) derives its two nitrogen atoms from two distinct sources within the urea cycle:
1. The first nitrogen enters the cycle as free Ammonia ($NH_3$), which combines with $CO_2$ to form carbamoyl phosphate.
2. The second nitrogen is donated by the amino acid Aspartate when it condenses with citrulline to form argininosuccinate.
Why other options are incorrect:
  • • Glutamate: While glutamate often provides the nitrogen to form ammonia (via deamination) or aspartate (via transamination), it is not the direct donor inside the urea cycle reactions.
  • • Glutamine: Glutamine is a non-toxic carrier of ammonia in the blood, but it must be converted to ammonia by glutaminase before the nitrogen can enter the cycle.
  • • Alanine: Alanine carries nitrogen from muscles to the liver, but its amino group must first be transferred to alpha-ketoglutarate to form glutamate, and then processed into ammonia or aspartate.
In carcinoid syndrome, the argentaffin tissue of the abdominal cavity overproduces:
  • a) Serotonin
  • b) Histamine
  • c) Tryptamine
  • d) Tyrosine
Explanation: In carcinoid syndrome, argentaffin (enterochromaffin) cells in carcinoid tumors overproduce serotonin (5-hydroxytryptamine). These tumors typically divert a large percentage of the body’s dietary tryptophan toward the synthesis of serotonin, which causes symptoms like flushing and diarrhea.
Why other options are incorrect:
  • • Histamine: While histamine can be produced by some gastric carcinoids, serotonin is the classic and primary substance overproduced in the typical syndrome.
  • • Tryptamine: This is a trace amine found in the brain and certain foods, but it is not the primary hormone overproduced in carcinoid syndrome.
  • • Tyrosine: This is an amino acid used as a building block for proteins and catecholamines (like dopamine), not a product overproduced by argentaffin tissue in this syndrome.
Tryptophan could be considered as a precursor of:
  • a) Melatonin
  • b) Thyroid hormones
  • c) Melanin
  • d) Epinephrine
Explanation: Tryptophan is the precursor for Melatonin. The pathway involves the conversion of Tryptophan to Serotonin, which is then converted into Melatonin (the sleep-regulating hormone) primarily in the pineal gland. Tryptophan is also a precursor for Niacin (Vitamin B3).
Why other options are incorrect:
  • • Thyroid hormones: These (T3 and T4) are synthesized from the amino acid Tyrosine.
  • • Melanin: This pigment is synthesized from Tyrosine via the action of the enzyme tyrosinase.
  • • Epinephrine: Also known as adrenaline, it is a catecholamine synthesized from Tyrosine (via DOPA, Dopamine, and Norepinephrine).
Conversion of tyrosine to dihydroxyphenylalanine (DOPA) is catalysed by tyrosine hydroxylase which requires:
  • a) NAD
  • b) FAD
  • c) ATP
  • d) Tetrahydrobiopterin
Explanation: Tyrosine hydroxylase is the rate-limiting enzyme in the synthesis of catecholamines. It requires Tetrahydrobiopterin (BH4), molecular oxygen, and ferrous iron ($Fe^{2+}$) as cofactors to convert Tyrosine into DOPA.
Why other options are incorrect:
  • • NAD / FAD: These are coenzymes involved in oxidation-reduction reactions in the electron transport chain and various metabolic pathways, but they do not serve as the specific reductant for aromatic amino acid hydroxylases.
  • • ATP: While many enzymatic reactions require ATP for energy, the hydroxylation of tyrosine is an oxidation reaction specifically dependent on the electron-donating properties of BH4.
The rate limiting step in the biosynthesis of catecholamines is:
  • a) Decarboxylation of dihydroxyphenylalanine
  • b) Hydroxylation of phenylalanine
  • c) Hydroxylation of tyrosine
  • d) Oxidation of dopamine
Explanation: The hydroxylation of tyrosine to form DOPA, catalyzed by the enzyme tyrosine hydroxylase, is the rate-limiting step in the synthesis of catecholamines (Dopamine, Norepinephrine, and Epinephrine). This enzyme is highly regulated by feedback inhibition from the end products.
Why other options are incorrect:
  • • Decarboxylation of DOPA: This step is catalyzed by DOPA decarboxylase. It occurs very rapidly and is not a regulatory or rate-limiting point in the pathway.
  • • Hydroxylation of phenylalanine: This step converts phenylalanine to tyrosine. While essential, the specific commitment to catecholamine synthesis begins with the next step (tyrosine to DOPA).
  • • Oxidation of dopamine: Dopamine is converted to Norepinephrine via hydroxylation (not oxidation) by dopamine $\beta$-hydroxylase. This is a subsequent step and not the primary rate-limiting one.
The enzyme dopamine β-oxidase (dopamine β-hydroxylase) which catalyses conversion of dopamine to norepinephrine requires:
  • a) Vitamin A
  • b) Vitamin C
  • c) Vitamin E
  • d) Vitamin B12
Explanation: The enzyme Dopamine β-hydroxylase (also known as dopamine β-oxidase) requires Vitamin C (Ascorbic acid) and copper ($Cu^{2+}$) as cofactors. Vitamin C acts as an electron donor in the reaction that converts dopamine into norepinephrine.
Why other options are incorrect:
  • • Vitamin A: Primarily involved in vision (retinal) and gene transcription; it does not play a role in catecholamine synthesis.
  • • Vitamin E: Functions as a lipid-soluble antioxidant to protect cell membranes from oxidative damage but is not a cofactor for this specific enzyme.
  • • Vitamin B12: Essential for DNA synthesis and the metabolism of amino acids and fatty acids (specifically methylmalonyl-CoA mutase and methionine synthase), but not for the hydroxylation of dopamine.
In humans, the sulphur of methionine and cysteine is excreted mainly as:
  • a) Ethereal sulphate
  • b) Inorganic sulphate
  • c) Sulphites
  • d) Thioorganic compound
Explanation: The sulphur atoms in the amino acids methionine and cysteine undergo oxidation during metabolism. In humans, the vast majority (approximately 80%) of this sulphur is excreted in the urine as inorganic sulphate ($SO_4^{2-}$).
Why other options are incorrect:
  • • Ethereal sulphate: These are organic sulphate esters (like indoxyl sulphate) formed in the liver to detoxify phenols; they represent only about 5-10% of total urinary sulphur.
  • • Sulphites: These are highly reactive metabolic intermediates that are normally rapidly converted to inorganic sulphates by the enzyme sulphite oxidase.
  • • Thioorganic compound: Also known as “neutral sulphur” (e.g., taurine, cystine, thiosulphate), these represent only a small fraction of the total sulphur excreted.
Small amount of urinary oxalates is contributed by the amino acid:
  • a) Glycine
  • b) Tyrosine
  • c) Alanine
  • d) Serine
Explanation: Glycine is a major metabolic precursor of oxalate. Glycine can be converted to glyoxylate, which is subsequently oxidized to oxalate by enzymes such as glycolate oxidase or xanthine oxidase. High levels of this process can contribute to the formation of calcium oxalate kidney stones.
Why other options are incorrect:
  • • Tyrosine: This is an aromatic amino acid used primarily for the synthesis of proteins, catecholamines, and melanin, not oxalate.
  • • Alanine: Alanine is mainly involved in transamination reactions and the glucose-alanine cycle for energy transport.
  • • Serine: While serine can be converted to glycine, it is glycine itself that is the direct intermediate for glyoxylate and oxalate production.

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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