4801 to 4850 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 4801 to 4850
Pyruvate kinase requires __ ions for maximum activity.
- a) Na+
- b) K+
- c) Ca2+
- d) Mg2+
Explanation:
Pyruvate kinase is one of the few enzymes that requires a monovalent cation for its activity. While it also requires divalent cations (like Mg2+ or Mn2+), Potassium (K+) is specifically required for maximum catalytic efficiency and to maintain the active conformation of the enzyme.
Why other options are incorrect:
Why other options are incorrect:
- • Na+: Sodium ions are generally not effective activators for pyruvate kinase and can sometimes act as a competitive inhibitor against K+.
- • Ca2+: Calcium ions typically act as an inhibitor of pyruvate kinase by competing with the required Mg2+ ions.
- • Mg2+: While Mg2+ is a necessary divalent cofactor, K+ is the specific monovalent ion requirement often highlighted for achieving “maximum” activity in biochemical contexts.
ATP is ‘wasted’ in Rapoport-Lueberring cycle in RBCs as otherwise it will inhibit:
- a) Phosphoglucomutase
- b) Phosphohexo isomerase
- c) Phosphofructo kinase
- d) Phosphoenol pyruvate carboxy kinase
Explanation:
In Red Blood Cells (RBCs), the Rapoport-Lueberring cycle bypasses the phosphoglycerate kinase step of glycolysis to produce 2,3-Bisphosphoglycerate (2,3-BPG). This “wastes” the production of ATP. This is necessary because RBCs have low energy requirements; if ATP levels become too high, ATP acts as a potent allosteric inhibitor of Phosphofructokinase (PFK-1), which would shut down glycolysis entirely.
Why other options are incorrect:
Why other options are incorrect:
- • Phosphoglucomutase: This enzyme is involved in glycogen metabolism (converting Glucose-1-P to Glucose-6-P) and is not the primary target of ATP inhibition in RBC glycolysis.
- • Phosphohexo isomerase: This enzyme catalyzes a reversible step in glycolysis and is not a major regulatory or rate-limiting step inhibited by ATP.
- • PEPCK: This is a key enzyme in gluconeogenesis (making glucose). RBCs do not perform gluconeogenesis; they only perform glycolysis.
The following co-enzyme is needed for the oxidative decarboxylation of ketoacids:
- a) NADP+
- b) TPP
- c) Folate coenzyme
- d) Biotin coenzyme
Explanation:
TPP (Thiamine Pyrophosphate), derived from Vitamin B1, is an essential co-enzyme for the oxidative decarboxylation of alpha-ketoacids (such as pyruvate and alpha-ketoglutarate). It works within multi-enzyme complexes to facilitate the removal of CO2 and the transfer of the remaining acyl group.
Why other options are incorrect:
Why other options are incorrect:
- • NADP+: Primarily functions as a reducing agent in anabolic (biosynthetic) pathways and the HMP shunt, rather than oxidative decarboxylation.
- • Folate coenzyme: Functions in the transfer of one-carbon units (like methyl or formyl groups) in the synthesis of amino acids and nucleic acids.
- • Biotin coenzyme: Biotin is required for carboxylation reactions (adding CO2), such as the conversion of pyruvate to oxaloacetate, which is the opposite of decarboxylation.
Synthesis of Glucose from amino acids is termed as:
- a) Glycolysis
- b) Gluconeogenesis
- c) Glycogenesis
- d) Lipogenesis
Explanation:
Gluconeogenesis is the metabolic process by which organisms produce glucose from non-carbohydrate precursors, including glucogenic amino acids, lactate, and glycerol. This occurs mainly in the liver and kidneys during periods of fasting or intense exercise.
Why other options are incorrect:
Why other options are incorrect:
- • Glycolysis: This is the breakdown of glucose to produce energy (pyruvate/lactate), which is the opposite of synthesis.
- • Glycogenesis: This is the process of synthesizing glycogen from glucose for storage, not the synthesis of glucose itself.
- • Lipogenesis: This is the metabolic process of converting acetyl-CoA into fatty acids or lipids.
The following examples are important heteropolysaccharides except:
- a) Amylopectin
- b) Heparin
- c) Peptidoglycan
- d) Hyaluronic acid
Explanation:
Amylopectin is a homopolysaccharide because it is composed entirely of a single type of monosaccharide unit: D-glucose. It is one of the two components of starch. Heteropolysaccharides, on the other hand, contain two or more different types of monosaccharides.
Why other options are incorrect:
Why other options are incorrect:
- • Heparin: A heteropolysaccharide (specifically a glycosaminoglycan) composed of repeating units of glucuronic acid and glucosamine.
- • Peptidoglycan: A heteropolysaccharide found in bacterial cell walls, consisting of alternating units of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM).
- • Hyaluronic acid: A heteropolysaccharide made of repeating disaccharide units of D-glucuronic acid and N-acetyl-D-glucosamine.
Which of the following features are common to monosaccharides?
- a) Contain asymmetric centres
- b) Are of 2 types – aldoses and ketoses
- c) Tend to exist as ring structures in solution
- d) Include glucose, galactose and raffinose
Explanation:
Monosaccharides are chemically defined as polyhydroxy aldehydes or polyhydroxy ketones. Therefore, the most universal classification common to all monosaccharides is that they are either aldoses (containing an aldehyde group) or ketoses (containing a ketone group).
Why other options are incorrect:
Why other options are incorrect:
- • Contain asymmetric centres: While most monosaccharides have them, dihydroxyacetone (a triose ketose) is a monosaccharide that has no asymmetric (chiral) carbon.
- • Tend to exist as ring structures: This is common for pentoses (5C) and hexoses (6C), but smaller monosaccharides like trioses (3C) and tetroses (4C) do not form stable ring structures in solution.
- • Include glucose, galactose and raffinose: Glucose and galactose are monosaccharides, but raffinose is a trisaccharide (composed of galactose, glucose, and fructose).
Polysaccharides
- a) Contain many monosaccharide units which may or may not be of the same kind
- b) Function mainly as storage or structural compounds
- c) Are present in large amounts in connective tissue
- d) All of these
Explanation:
Polysaccharides are carbohydrates composed of many monosaccharide units. They may be homopolysaccharides, containing the same type of monosaccharide, or heteropolysaccharides, containing different types. They mainly serve storage and structural functions and are important components of connective tissue. Therefore, all of the statements are correct.
Why other options are incorrect:
Why other options are incorrect:
- • Option A: Correctly describes the monosaccharide composition of polysaccharides.
- • Option B: Correctly describes their major storage and structural roles.
- • Option C: Polysaccharides such as glycosaminoglycans are important components of connective tissue.
The absorption of glucose in the digestive tract:
- a) Occurs in the small intestine
- b) Is stimulated by the hormone Glucagon
- c) Occurs more rapidly than the absorption of any other sugar
- d) Is impaired in cases of diabetes mellitus
Explanation:
The primary site for the absorption of monosaccharides, including glucose, is the small intestine (specifically the duodenum and upper jejunum). Glucose is absorbed via active transport using the Sodium-Glucose Co-transporter 1 (SGLT1) and facilitated diffusion via GLUT2.
Why other options are incorrect:
Why other options are incorrect:
- • Glucagon: This hormone increases blood glucose levels by acting on the liver (glycogenolysis); it does not stimulate the absorption of glucose from the digestive tract.
- • Rate of absorption: While glucose is absorbed very quickly, galactose is actually absorbed at a rate slightly faster than or equal to glucose.
- • Diabetes Mellitus: In diabetes, the problem lies in the uptake of glucose from the blood into peripheral tissues (like muscle or fat) or insulin production, not the initial absorption of glucose from the gut.
UDP-Glucose is converted to UDP-Glucuronic acid by:
- a) ATP
- b) GTP
- c) NADP+
- d) NAD+
Explanation:
The enzyme UDP-glucose dehydrogenase catalyzes the conversion of UDP-glucose to UDP-glucuronic acid. This is a two-step oxidation of the primary alcohol group at carbon 6 of the glucose moiety, and it specifically requires NAD+ as the electron acceptor (it produces NADH).
Why other options are incorrect:
Why other options are incorrect:
- • ATP / GTP: These are high-energy phosphate donors. While energy is needed to initially “activate” glucose to UDP-glucose, the conversion to glucuronic acid is an oxidation-reduction reaction, not a phosphorylation.
- • NADP+: Although similar to NAD+, this specific dehydrogenase enzyme in the uronic acid pathway is highly specific for NAD+. NADP+ is typically reserved for biosynthetic reductions or the HMP shunt.
The enzymes involved in Phosphorylation of glucose to glucose 6- phosphate are:
- a) Hexokinase
- b) Glucokinase
- c) Phosphofructokinase
- d) Both (Hexokinase) and (Glucokinase)
Explanation:
The first step of glycolysis is the phosphorylation of glucose to glucose 6-phosphate. This is catalyzed by two different enzymes depending on the tissue:
Why other options are incorrect:
- Hexokinase: Found in most extrahepatic tissues. It has a low Km (high affinity for glucose) and is inhibited by its product, glucose 6-phosphate.
- Glucokinase (Hexokinase IV): Found specifically in the liver and pancreatic β-cells. It has a high Km (low affinity) and is active only when blood glucose levels are high.
Why other options are incorrect:
- • Phosphofructokinase (PFK): This enzyme catalyzes the third step of glycolysis (converting Fructose 6-phosphate to Fructose 1,6-bisphosphate), which is the rate-limiting step of the pathway.
In conversion of Lactic acid to Glucose, three reactions of Glycolytic pathway are circumvented, which of the following enzymes do not participate?
- a) Pyruvate Carboxylase
- b) Phosphoenol pyruvate carboxy kinase
- c) Pyruvate kinase
- d) Glucose-6-phosphatase
Explanation:
During conversion of lactic acid to glucose by gluconeogenesis, the three irreversible steps of glycolysis are bypassed by different enzymes. Pyruvate kinase is bypassed and does not participate in gluconeogenesis. Instead, pyruvate carboxylase and phosphoenolpyruvate carboxykinase convert pyruvate toward phosphoenolpyruvate, while glucose-6-phosphatase catalyzes the final step of glucose formation.
Why other options are incorrect:
Why other options are incorrect:
- • Pyruvate Carboxylase: Converts pyruvate to oxaloacetate during gluconeogenesis.
- • Phosphoenol pyruvate carboxy kinase: Converts oxaloacetate to phosphoenolpyruvate.
- • Glucose-6-phosphatase: Catalyzes the final bypass step, producing free glucose from glucose-6-phosphate.
The normal resting state of humans, most of the blood glucose burnt as “fuel” is consumed by:
- a) Liver
- b) Brain
- c) Kidneys
- d) Adipose tissue
Explanation:
In a resting state, the Brain is the largest consumer of blood glucose. It accounts for approximately 20% of the body’s total oxygen consumption and uses about 120 grams of glucose daily. Unlike other tissues, the brain has a very limited ability to store fuel and relies on a continuous supply of glucose from the blood.
Why other options are incorrect:
Why other options are incorrect:
- • Liver: While the liver is central to glucose metabolism, its primary role is to regulate blood glucose (via glycogenolysis and gluconeogenesis) rather than acting as the primary consumer of it for its own energy at rest.
- • Kidneys: While the kidneys require energy for filtration and reabsorption, their total glucose consumption is significantly lower than that of the brain.
- • Adipose tissue: Fat cells use glucose mainly to produce glycerol-3-phosphate for fat storage (triglyceride synthesis), but they do not “burn” large amounts of glucose for fuel compared to the metabolic demands of the nervous system.
Which of the following compound is a positive allosteric modifier of the enzyme pyruvate carboxylase?
- a) Biotin
- b) Acetyl CoA
- c) Oxaloacetate
- d) ATP
Explanation:
Acetyl-CoA is a classic positive allosteric activator of pyruvate carboxylase. High levels of Acetyl-CoA signal that the TCA cycle is “saturated” or that there is an abundance of metabolic fuel (from fatty acid oxidation). This triggers the enzyme to convert pyruvate into oxaloacetate to either replenish the TCA cycle or initiate gluconeogenesis.
Why other options are incorrect:
Why other options are incorrect:
- • Biotin: This is a prosthetic group (coenzyme) that is covalently attached to the enzyme. It is essential for the reaction mechanism but is not considered an allosteric modifier.
- • Oxaloacetate: This is the product of the reaction. In metabolic pathways, high concentrations of products usually lead to feedback inhibition, not positive activation.
- • ATP: While the reaction requires ATP as a substrate (energy source) to function, Acetyl-CoA is the specific regulatory molecule that acts as the allosteric “switch” for the enzyme.
A specific inhibitor for succinate dehydrogenase is:
- a) Arsinite
- b) Melouate
- c) Citrate
- d) Cyanide
Explanation:
Malonate (often transcribed as Melouate in some texts) is a classic competitive inhibitor of succinate dehydrogenase. Because its structure is very similar to succinate (both are dicarboxylic acids), it competes for the enzyme’s active site but cannot be oxidized to fumarate.
Why other options are incorrect:
Why other options are incorrect:
- • Arsinite: Primarily inhibits enzymes that require lipoic acid as a cofactor, such as the Pyruvate Dehydrogenase complex.
- • Citrate: This is a substrate in the TCA cycle. While it can allosterically inhibit Phosphofructokinase (PFK) in glycolysis, it is not a specific inhibitor of succinate dehydrogenase.
- • Cyanide: A potent inhibitor of the Electron Transport Chain, but it specifically targets Cytochrome c oxidase (Complex IV), not succinate dehydrogenase (Complex II).
Most of the metabolic pathways are either anabolic or catabolic. Which of the following pathways is considered as “amphibolic” in nature?
- a) Glycogenesis
- b) Glycolytic pathway
- c) Lipolysis
- d) TCA cycle
Explanation:
An amphibolic pathway is one that functions in both catabolism (breakdown) and anabolism (synthesis). The TCA cycle (Krebs cycle) is the best example because it oxidizes Acetyl-CoA to CO2 and water to produce energy (catabolism), but also provides key intermediates for the synthesis of glucose, amino acids, and heme (anabolism).
Why other options are incorrect:
Why other options are incorrect:
- • Glycogenesis: This is a purely anabolic pathway (the synthesis of glycogen from glucose).
- • Glycolytic pathway: This is primarily a catabolic pathway (the breakdown of glucose to pyruvate for energy production).
- • Lipolysis: This is a purely catabolic process (the breakdown of fats/triacylglycerols into fatty acids and glycerol).
Transketolase activity is affected in:
- a) Biotin deficiency
- b) Pyridoxine deficiency
- c) PABA deficiency
- d) Thiamine deficiency
Explanation:
Transketolase is a key enzyme in the Hexose Monophosphate (HMP) shunt (Pentose Phosphate Pathway). It requires Thiamine Pyrophosphate (TPP), which is the active form of Vitamin B1 (Thiamine), as a coenzyme. Measuring erythrocyte transketolase activity is actually a standard clinical laboratory test used to diagnose thiamine deficiency.
Why other options are incorrect:
Why other options are incorrect:
- • Biotin deficiency: Biotin is a cofactor for carboxylation reactions (like Pyruvate carboxylase), not for transketolase.
- • Pyridoxine (B6) deficiency: Vitamin B6 is essential for transamination and decarboxylation of amino acids.
- • PABA deficiency: Para-aminobenzoic acid is a precursor for folic acid synthesis in bacteria and does not act as a cofactor for human transketolase.
The following metabolic abnormalities occur in Diabetes mellitus except:
- a) Increased plasma FFA
- b) Increased pyruvate carboxylase activity
- c) Decreased lipogenesis
- d) Decreased gluconeogenesis
Explanation:
In Diabetes Mellitus, gluconeogenesis is increased, not decreased. Due to insulin deficiency or resistance, the liver overproduces glucose from non-carbohydrate sources to compensate for the perceived lack of energy in cells, contributing to hyperglycemia.
Why other options are incorrect:
Why other options are incorrect:
- • Increased plasma FFA: Insulin deficiency leads to increased lipolysis in adipose tissue, releasing high amounts of Free Fatty Acids (FFA) into the blood.
- • Increased pyruvate carboxylase activity: Pyruvate carboxylase is a key regulatory enzyme in gluconeogenesis; its activity increases in diabetes to drive glucose production.
- • Decreased lipogenesis: Since insulin is required to stimulate fat synthesis, its absence or lack of effect leads to a significant decrease in the storage of new fats.
A substance that is not an intermediate in the formation of D-glucuronic acid from glucose is:
- a) Glucose-1-p
- b) 6-Phosphogluconate
- c) Glucose-6-p
- d) UDP-Glucose
Explanation:
The formation of D-glucuronic acid occurs via the Uronic Acid Pathway. 6-Phosphogluconate is not part of this pathway; instead, it is a key intermediate in the Pentose Phosphate Pathway (Hexose Monophosphate Shunt).
Why other options are incorrect:
Why other options are incorrect:
- • Glucose-6-p: The first step where Glucose is phosphorylated to enter the metabolic pathway.
- • Glucose-1-p: Formed from Glucose-6-P by the enzyme phosphoglucomutase before it is converted to UDP-Glucose.
- • UDP-Glucose: A critical intermediate that is oxidized by UDP-glucose dehydrogenase to eventually form Glucuronic acid.
The hydrolysis of Glucose-6-P is catalysed by a phosphatase that is not formed in which of the following?
- a) Liver
- b) Kidney
- c) Muscle
- d) Small intestine
Explanation:
The enzyme Glucose-6-phosphatase is responsible for converting Glucose-6-Phosphate into free glucose. This enzyme is present in the liver, kidneys, and small intestinal mucosa, allowing these organs to release glucose into the blood. However, it is absent in muscle.
Why other options are incorrect:
Why other options are incorrect:
- • Liver: The primary site for this enzyme; it plays a major role in maintaining blood glucose levels via glycogenolysis and gluconeogenesis.
- • Kidney: Contains the enzyme and contributes significantly to blood glucose levels, especially during prolonged fasting.
- • Small intestine: Contains the enzyme to facilitate the transport of glucose into the bloodstream during the post-absorptive state.
- • Note on Muscle: Because muscles lack this enzyme, they cannot release glucose into the blood; instead, the Glucose-6-P is used internally for energy (glycolysis).
An essential for converting Glucose to Glycogen in Liver is:
- a) Lactic acid
- b) GTP
- c) CTP
- d) UTP
Explanation:
UTP (Uridine Triphosphate) is essential for glycogenesis. It reacts with Glucose-1-phosphate to form UDP-glucose, which is the active glucosyl donor used by the enzyme glycogen synthase to extend the glycogen chain.
Why other options are incorrect:
Why other options are incorrect:
- • Lactic acid: This is a byproduct of anaerobic glycolysis, not a required nucleotide or cofactor for building glycogen.
- • GTP: Guanosine triphosphate is primarily involved in protein synthesis and the gluconeogenesis step (PEPCK reaction), but not directly in glycogen synthesis.
- • CTP: Cytidine triphosphate is mainly used in the biosynthesis of phospholipids.
Which of the following is a substrate for aldolase activity in Glycolytic pathway?
- a) Glyceraldehyde-3-p
- b) Glucose-6-p
- c) Fructose-6-p
- d) Fructose 1, 6-bisphosphate
Explanation:
In the glycolytic pathway, the enzyme Aldolase (specifically Aldolase A) catalyzes the cleavage of Fructose 1,6-bisphosphate into two three-carbon molecules: Glyceraldehyde-3-phosphate (GAP) and Dihydroxyacetone phosphate (DHAP).
Why other options are incorrect:
Why other options are incorrect:
- • Glyceraldehyde-3-p: This is a product of the aldolase reaction, not the substrate.
- • Glucose-6-p: This is the substrate for the enzyme Phosphohexose Isomerase.
- • Fructose-6-p: This is the substrate for Phosphofructokinase-1 (PFK-1), which adds a phosphate group to form Fructose 1,6-bisphosphate.
The ratio that approximates the number of net molecule of ATP formed per mole of Glucose oxidized in presence of O2 to the net number formed in absence of O2 is:
- a) 4 : 1
- b) 10 : 2
- c) 12 : 1
- d) 18 : 1
Explanation:
This ratio is calculated based on the net ATP yield of aerobic vs. anaerobic respiration:
Why other options are incorrect:
- Aerobic Respiration: Complete oxidation of 1 mole of glucose yields approximately 36 to 38 ATP (depending on the shuttle system used).
- Anaerobic Respiration: Conversion of 1 mole of glucose to lactate yields a net of only 2 ATP.
Why other options are incorrect:
- • 4 : 1: This would imply only 8 ATP are formed aerobically, which is far too low.
- • 10 : 2: This simplifies to 5:1, significantly underestimating the efficiency of the Mitochondria/TCA cycle.
- • 12 : 1: This would imply 24 ATP are formed aerobically; while higher than other options, it still doesn’t account for the full yield of the Electron Transport Chain.
The “Primaquin sensitivity” type of haemolytic anaemia has been found to relate to reduced R.B.C activity of which enzyme?
- a) Pyruvate kinase deficiency
- b) Glucose-6-phosphatase deficiency
- c) Glucose-6-p dehydrogenase deficiency
- d) Hexokinase deficiency
Explanation:
Glucose-6-phosphate dehydrogenase (G6PD) deficiency is an X-linked genetic disorder. G6PD is critical for producing NADPH, which maintains reduced glutathione in red blood cells. Reduced glutathione protects RBCs from oxidative damage. When patients with this deficiency take oxidative drugs like Primaquin, the RBCs undergo hemolysis.
Why other options are incorrect:
Why other options are incorrect:
- • Pyruvate kinase deficiency: While this also causes hemolytic anemia, it is due to ATP depletion in the glycolytic pathway and is not specifically triggered by Primaquin.
- • Glucose-6-phosphatase deficiency: This leads to Von Gierke disease (Glycogen Storage Disease Type I), affecting the liver and kidneys, not red blood cell stability.
- • Hexokinase deficiency: A very rare cause of hereditary non-spherocytic hemolytic anemia, but it is not the classic “Primaquin-sensitive” type.
Which of the following hormones is not involved in carbohydrate metabolism?
- a) Cortisol
- b) ACTH
- c) Glucagon
- d) Vasopressin
Explanation:
Vasopressin (also known as Antidiuretic Hormone or ADH) is primarily responsible for regulating water retention by the kidneys and blood pressure. It does not play a direct role in carbohydrate metabolism.
Why other options are incorrect:
Why other options are incorrect:
- • Cortisol: A glucocorticoid that increases blood glucose levels by stimulating gluconeogenesis in the liver.
- • ACTH: Adrenocorticotropic hormone stimulates the adrenal cortex to release cortisol, thus indirectly impacting carbohydrate metabolism.
- • Glucagon: A primary hormone that raises blood glucose levels by promoting glycogenolysis and gluconeogenesis.
Dehydrogenases involved in HMP shunt are specific for:
- a) NADP+
- b) NAD+
- c) FAD
- d) FMN
Explanation:
The HMP Shunt (Hexose Monophosphate Shunt or Pentose Phosphate Pathway) is the primary metabolic pathway that generates NADPH. The two oxidative enzymes in this pathway—Glucose-6-phosphate dehydrogenase (G6PD) and 6-phosphogluconate dehydrogenase—specifically use NADP+ as their electron acceptor.
Why other options are incorrect:
Why other options are incorrect:
- • NAD+: This coenzyme is typically used in catabolic reactions like Glycolysis and the TCA cycle to generate NADH for energy production (ATP).
- • FAD: Flavin Adenine Dinucleotide is used in different redox reactions, such as those catalyzed by succinate dehydrogenase in the TCA cycle.
- • FMN: Flavin Mononucleotide is primarily involved in the Electron Transport Chain (Complex I) and does not serve as a coenzyme for HMP shunt dehydrogenases.
Which of the following enzymes in Glycolytic pathway is inhibited by fluoride?
- a) Glyceraldehyde-3-p dehydrogenase
- b) Phosphoglycerate kinase
- c) Pyruvate kinase
- d) Enolase
Explanation:
The enzyme Enolase is strongly inhibited by fluoride. Fluoride ions combine with magnesium and phosphate to form a fluorophosphate-magnesium complex, which displaces the magnesium required by the enzyme for activity. This is why sodium fluoride is used in blood collection tubes to prevent glycolysis before glucose testing.
Why other options are incorrect:
Why other options are incorrect:
- • Glyceraldehyde-3-p dehydrogenase: This enzyme is typically inhibited by iodoacetate or arsenate, not fluoride.
- • Phosphoglycerate kinase: This is a substrate-level phosphorylation step that is not the target of fluoride inhibition.
- • Pyruvate kinase: This is the final regulatory step of glycolysis; it is inhibited by high ATP and Alanine, but not by fluoride.
Out of 24 mols of ATP formed in TCA cycle, 2 molecules of ATP can be formed at “substrate level” by which of the following reaction?
- a) Citric acid → Isocitric acid
- b) Isocitrate → Oxaloacetate
- c) Succinic acid → Fumarate
- d) Succinyl-CoA → Succinic acid
Explanation:
Substrate-level phosphorylation refers to the direct formation of ATP (or GTP) without the electron transport chain. In the TCA cycle, the conversion of Succinyl-CoA to Succinic acid (catalyzed by Succinyl-CoA synthetase) is the only step that produces a high-energy phosphate bond (GTP/ATP) directly.
Since one glucose molecule produces two Acetyl-CoA molecules, this reaction occurs twice, resulting in 2 molecules of ATP per mole of glucose.
Why other options are incorrect:
Since one glucose molecule produces two Acetyl-CoA molecules, this reaction occurs twice, resulting in 2 molecules of ATP per mole of glucose.
Why other options are incorrect:
- • Citric acid → Isocitric acid: This is an isomerization reaction catalyzed by aconitase; no energy is produced.
- • Isocitrate → Oxaloacetate: This is a multi-step process involving the formation of NADH, which produces ATP only via oxidative phosphorylation (ETC), not substrate level.
- • Succinic acid → Fumarate: This reaction produces FADH₂, which subsequently generates ATP via the electron transport chain (oxidative phosphorylation).
Which of the following statements regarding T.C.A cycle is true?
- a) It is an anaerobic process
- b) It occurs in cytosol
- c) It contains no intermediates for Gluconeogenesis
- d) It is amphibolic in nature
Explanation:
The TCA cycle is described as amphibolic because it serves a dual purpose. It is catabolic (it breaks down Acetyl-CoA to produce energy/ATP) and anabolic (its intermediates are used as precursors for the synthesis of glucose, amino acids, fatty acids, and heme).
Why other options are incorrect:
Why other options are incorrect:
- • It is an anaerobic process: Incorrect. The TCA cycle is strictly aerobic; it requires oxygen indirectly to regenerate NAD+ and FAD through the Electron Transport Chain.
- • It occurs in cytosol: Incorrect. The TCA cycle enzymes are located in the mitochondrial matrix (except for succinate dehydrogenase, which is on the inner mitochondrial membrane).
- • It contains no intermediates for Gluconeogenesis: Incorrect. Intermediates like oxaloacetate, malate, and α-ketoglutarate are essential starting points for gluconeogenesis and amino acid synthesis.
An allosteric enzyme responsible for controlling the rate of T.C.A cycle is:
- a) Malate dehydrogenase
- b) Isocitrate dehydrogenase
- c) Fumarase
- d) Aconitase
Explanation:
Isocitrate dehydrogenase is considered the primary rate-limiting and allosteric enzyme of the TCA cycle. It is positively regulated (activated) by ADP and Ca²⁺, and negatively regulated (inhibited) by ATP and NADH.
Why other options are incorrect:
Why other options are incorrect:
- • Malate dehydrogenase: While part of the cycle, it catalyzes a reaction that is close to equilibrium and is not a major allosteric control point.
- • Fumarase: This enzyme catalyzes the reversible hydration of fumarate to malate and does not have a regulatory role in the cycle’s speed.
- • Aconitase: This enzyme isomerizes citrate to isocitrate. It is an equilibrium enzyme and is not involved in the allosteric regulation of the cycle’s rate.
The glycolysis is regulated by
- a) Hexokinase
- b) Phosphofructokinase
- c) Pyruvate kinase
- d) All of these
Explanation:
Glycolysis is regulated at three major irreversible steps catalyzed by hexokinase, phosphofructokinase-1 (PFK-1), and pyruvate kinase. These enzymes control the rate of glycolytic flux in response to cellular energy requirements.
Why other options are incorrect:
Why other options are incorrect:
- • Hexokinase: Catalyzes the first regulatory step of glycolysis.
- • Phosphofructokinase: PFK-1 catalyzes the major rate-limiting regulatory step of glycolysis.
- • Pyruvate kinase: Catalyzes the final irreversible, regulated step of glycolysis.
How many ATP molecules will be required for conversion of 2-molecules of Lactic acid to Glucose?
- a) 2
- b) 4
- c) 8
- d) 6
Explanation:
The conversion of 2 molecules of lactate to 1 molecule of glucose (Gluconeogenesis) requires 6 ATP molecules. For each molecule of lactate converted to pyruvate and then to glucose, 3 ATP equivalents are consumed: 1 ATP at Pyruvate Carboxylase, 1 GTP (ATP equivalent) at PEP Carboxykinase, and 1 ATP at Phosphoglycerate Kinase. Since 2 lactates are needed for 1 glucose, the total is $3 \times 2 = 6$.
Why other options are incorrect:
Why other options are incorrect:
- • 2 ATP: This is the net gain during Glycolysis, but the reverse process (Gluconeogenesis) requires significantly more energy to bypass irreversible steps.
- • 4 ATP: This value does not account for all the phosphorylation steps required in the gluconeogenic pathway.
- • 8 ATP: This is an overestimation of the chemical energy required for this specific metabolic conversion.
Which of the following enzyme is not involved in HMP shunt?
- a) Glyceraldehyde-3-P dehydrogenase
- b) Glucose-6-P dehydrogenase
- c) Transketolase
- d) Phosphogluconate dehydrogenase
Explanation:
Glyceraldehyde-3-phosphate dehydrogenase is an enzyme of glycolysis and is not involved in the HMP shunt (pentose phosphate pathway). The HMP shunt uses enzymes such as glucose-6-phosphate dehydrogenase, phosphogluconate dehydrogenase, and transketolase.
Why other options are incorrect:
Why other options are incorrect:
- • Glucose-6-P dehydrogenase: Catalyzes the first and rate-limiting step of the HMP shunt.
- • Transketolase: Participates in the non-oxidative phase of the HMP shunt.
- • Phosphogluconate dehydrogenase: Participates in the oxidative phase of the HMP shunt.
In presence of the following cofactor, pyruvate carboxylase converts pyruvate to oxaloacetate:
- a) ATP, Protein and CO2
- b) CO2 and ATP
- c) CO2
- d) Protein
Explanation:
Pyruvate carboxylase converts pyruvate to oxaloacetate in a reaction that requires ATP and CO2. The enzyme also uses biotin as its cofactor.
Why other options are incorrect:
Why other options are incorrect:
- • ATP, Protein and CO2: Protein is not a required cofactor or reactant in this reaction.
- • CO2: CO2 is required, but ATP is also necessary to provide energy for the carboxylation reaction.
- • Protein: Protein alone cannot provide the energy or carbon dioxide required for the reaction.
For conversion of oxaloacetate to phosphoenol pyruvate, high energy molecule is required in the form of
- a) GTP only
- b) ITP only
- c) GTP (or) ITP
- d) None of these
Explanation:
During gluconeogenesis, oxaloacetate is converted to phosphoenolpyruvate (PEP) by phosphoenolpyruvate carboxykinase (PEPCK). This reaction requires a high-energy nucleotide, usually GTP, and ITP can also serve as an alternative phosphate donor.
Why other options are incorrect:
Why other options are incorrect:
- • GTP only: GTP is the usual energy source, but ITP can also support the reaction.
- • ITP only: ITP can serve as a phosphate donor, but it is not the only high-energy molecule involved.
- • None of these: Incorrect because GTP or ITP can provide the required high-energy phosphate.
If the more negative standard reduction potential of a redox pair, the greater the tendency to:
- a) To lose electrons
- b) To gain electrons
- c) To lose/gain electrons
- d) To lose and gain electrons
Explanation:
The standard reduction potential measures the tendency of a species to be reduced (gain electrons). A more negative value indicates that the species is a stronger reducing agent, meaning it prefers to undergo oxidation and lose electrons.
Why other options are incorrect:
Why other options are incorrect:
- • To gain electrons: This occurs when the reduction potential is more positive (stronger oxidizing agents).
- • To lose/gain electrons: The standard reduction potential specifically indicates a direction of preference; a negative value specifically points toward losing electrons.
- • To lose and gain electrons: This describes a full redox reaction, whereas the potential of a single pair describes its specific tendency toward one half-reaction.
Electron transport and phosphorylation can be uncoupled by compounds that increase the permeability of the inner mitochondrial membrane to:
- a) Electrons
- b) Protons
- c) Uncouplers
- d) All of these
Explanation:
Uncouplers (like 2,4-DNP) work by carrying protons (H+) across the inner mitochondrial membrane, bypassing the ATP synthase (Complex V). This dissipates the proton gradient, meaning energy is released as heat instead of being used to synthesize ATP.
Why other options are incorrect:
Why other options are incorrect:
- • Electrons: Electrons travel through the Electron Transport Chain (ETC) complexes within the membrane; uncoupling specifically refers to the disruption of the chemiosmotic proton gradient.
- • Uncouplers: While uncouplers are the agents that cause the effect, the question asks what the membrane becomes permeable to. The uncoupling effect is specifically due to the leakage of protons.
- • All of these: Incorrect because the mechanism is specific to proton permeability and its effect on the electrochemical gradient.
The more positive the E0, the greater the tendency of the oxidant member of that pair to:
- a) Lose electrons
- b) Gain electrons
- c) Lose (or) gain electrons
- d) Lose and gain electrons
Explanation:
The standard reduction potential (E0) is a measure of the affinity of a substance for electrons. A more positive E0 indicates a higher affinity for electrons, meaning the oxidant member of the pair has a greater tendency to be reduced (gain electrons).
Why other options are incorrect:
Why other options are incorrect:
- • Lose electrons: This occurs when the E0 is more negative (stronger reducing agents/tendency for oxidation).
- • Lose (or) gain electrons: This is vague; the positive sign specifically dictates the direction towards gaining electrons.
- • Lose and gain electrons: A single reduction potential value describes a specific directional tendency (reduction), not both processes simultaneously.
The standard free energy of hydrolysis of the terminal phosphate group of ATP is:
- a) –7,300 cal/mol
- b) –8,300 cal/mol
- c) 10,000 cal/mol
- d) +7,300 cal/mol
Explanation:
The hydrolysis of ATP to ADP and inorganic phosphate (Pi) is an exergonic reaction. Under standard conditions (pH 7.0, 25°C), the standard free energy change ($\Delta G^\circ’$) is approximately –7,300 cal/mol (or –7.3 kcal/mol, which equals –30.5 kJ/mol).
Why other options are incorrect:
Why other options are incorrect:
- • –8,300 cal/mol: This value does not correspond to the standard biological constant for ATP hydrolysis.
- • 10,000 cal/mol: While the actual free energy ($\Delta G$) in a living cell can be higher (around -11 to -12 kcal/mol), the standard free energy ($\Delta G^\circ’$) is specifically defined as -7.3 kcal/mol.
- • +7,300 cal/mol: The positive sign would indicate an endergonic reaction (energy absorbing), but ATP hydrolysis releases energy (exergonic).
The transport of a pair of electrons from NADH to O2 via the electron transport chain produces:
- a) –52,580 cal
- b) –50,580 cal
- c) 21,900 cal
- d) +52,580 cal
Explanation:
The standard free energy change (ΔG°’) can be calculated using the formula ΔG°’ = –nFΔE₀’. For the transfer of 2 electrons from NADH to Oxygen, the potential difference (ΔE₀’) is 1.14V.
Calculation: –(2) × (23,062 cal/V) × (1.14 V) ≈ –52,580 cal. The negative sign indicates that energy is released (exergonic).
Why other options are incorrect:
Calculation: –(2) × (23,062 cal/V) × (1.14 V) ≈ –52,580 cal. The negative sign indicates that energy is released (exergonic).
Why other options are incorrect:
- • –50,580 cal: This value is mathematically incorrect based on the standard reduction potentials of NADH and Oxygen.
- • 21,900 cal: This represents the amount of energy captured in the form of 3 ATPs (3 × 7,300 cal), not the total energy produced by the electron transport.
- • +52,580 cal: A positive value would mean the reaction requires an input of energy (endergonic), but the oxidation of NADH is a highly spontaneous, energy-releasing process.
Sufficient energy required to produce 3 ATP from 3 ADP and 3 Pi is:
- a) –21,900 cal
- b) 29,900 cal
- c) 31,900 cal
- d) 39,900 cal
Explanation:
The standard free energy required to synthesize one mole of ATP from ADP and Pi is approximately +7,300 cal/mol (7.3 kcal/mol). Therefore, for 3 moles of ATP:
$3 \times 7,300 \text{ cal} = 21,900 \text{ cal}$.
Why other options are incorrect:
$3 \times 7,300 \text{ cal} = 21,900 \text{ cal}$.
Why other options are incorrect:
- • 29,900 cal / 31,900 cal: These values do not correspond to the standard calculation of $3 \times 7.3 \text{ kcal/mol}$.
- • 39,900 cal: While the actual energy required in a living cell (under non-standard conditions) is higher—often cited around 10–12 kcal per ATP—the standard textbook answer for “sufficient energy” is usually based on the $7,300$ constant.
The free energy change, ΔG:
- a) Is directly proportional to the standard free energy change, ΔG°
- b) Is equal to zero at equilibrium
- c) Can only be calculated when the reactants and products are present at 1mol/l concentrations
- d) Is equal to –RT ln Keq
Explanation:
By definition, at equilibrium, ΔG = 0. This means there is no net driving force for the reaction to proceed in either the forward or reverse direction.
Why other options are incorrect:
Why other options are incorrect:
- • Directly proportional to ΔG°: ΔG depends on both ΔG° and the actual concentrations of reactants and products (via the equation ΔG = ΔG° + RT ln Q), so it is not a simple direct proportion.
- • 1mol/l concentrations: When all reactants and products are at 1.0 M concentration, ΔG is equal to ΔG° (the standard free energy change). However, ΔG can be calculated at any concentration.
- • Equal to –RT ln Keq: This formula actually defines the standard free energy change (ΔG°), not the actual free energy change (ΔG).
Under standard conditions:
- a) The free energy change ΔG°, is equal to 0
- b) The standard free energy change ΔG, is equal to 0
- c) The free energy change ΔG, is equal to the standard free energy change ΔG°
- d) Keq is equal to 1
Explanation:
The actual free energy change is defined by the equation: $\Delta G = \Delta G^\circ + RT \ln Q$. Under standard conditions, all reactants and products are at 1.0 M concentration, making the reaction quotient ($Q$) equal to 1. Since $\ln(1) = 0$, the equation simplifies to $\Delta G = \Delta G^\circ$.
Why other options are incorrect:
Why other options are incorrect:
- • ΔG° is equal to 0: ΔG° is a fixed constant for a specific reaction; it is only zero if the equilibrium constant ($K_{eq}$) is exactly 1.
- • ΔG is equal to 0: This only occurs when a system has reached equilibrium, which is not the same as being under standard conditions.
- • Keq is equal to 1: $K_{eq}$ is determined by the nature of the chemical reaction, not by the starting concentrations (standard conditions).
An uncoupler of oxidative phosphorylation such as dinitrophenol:
- a) Inhibits electron transport and ATP synthesis
- b) Allow electron transport to proceed without ATP synthesis
- c) Inhibits electron transport without impairment of ATP synthesis
- d) Specially inhibits cytochrome b
Explanation:
Uncouplers like 2,4-dinitrophenol (DNP) “uncouple” the linkage between the Electron Transport Chain (ETC) and ATP synthase. They carry protons across the inner mitochondrial membrane, dissipating the proton gradient. As a result, the ETC continues to operate (and often speeds up), but the energy is released as heat instead of being captured as ATP.
Why other options are incorrect:
Why other options are incorrect:
- • Inhibits electron transport and ATP synthesis: This is the action of respiratory inhibitors (like Cyanide or Azide), not uncouplers.
- • Inhibits electron transport without impairment of ATP synthesis: This is impossible because ATP synthesis in mitochondria is dependent on the energy generated by electron transport.
- • Specially inhibits cytochrome b: This is the specific function of Antimycin A, a complex III inhibitor.
All of the following statements about the enzymic complex that carries out the synthesis of ATP during oxidative phosphorylation are correct except:
- a) It is located on the matrix side of the inner mitochondrial membrane
- b) It is inhibited by oligomycin
- c) It can exhibit ATPase activity
- d) It can bind molecular O2
Explanation:
The enzymic complex referred to is ATP Synthase (Complex V). Binding molecular oxygen ($O_2$) is the function of Cytochrome c Oxidase (Complex IV), not ATP Synthase. ATP Synthase’s primary role is to use the proton gradient to phosphorylate ADP into ATP.
Why other options are correct statements:
Why other options are correct statements:
- • Matrix side: The $F_1$ subunit of ATP synthase protrudes into the mitochondrial matrix where ATP synthesis occurs.
- • Oligomycin: This antibiotic specifically inhibits ATP synthase by blocking the $F_o$ subunit (the proton channel).
- • ATPase activity: Under certain experimental conditions or when isolated, the complex can work in reverse, hydrolyzing ATP (acting as an ATPase).
Glucokinase:
- a) Is widely distributed and occurs in most mammalian tissues
- b) Has a high Km for glucose and hence is important in the phosphorylation of glucose primarily after ingestion of a carbohydrate rich meal
- c) Is widely distributed in Prokaryotes
- d) None of these
Explanation:
Glucokinase (Hexokinase IV) has a high Km (low affinity) for glucose. This means it only functions at significant rates when blood glucose levels are elevated, such as after a meal, allowing the liver to remove excess glucose from the blood.
Why other options are incorrect:
Why other options are incorrect:
- • Option a: Hexokinase (Types I, II, and III) is widely distributed in most tissues, whereas Glucokinase is primarily found in the liver and pancreatic beta cells.
- • Option c: Glucokinase is a characteristic enzyme of specific eukaryotic tissues (like the mammalian liver); Prokaryotes use different pathways for glucose phosphorylation.
The reaction catalysed by phosphofructokinase:
- a) Is activated by high concentrations of ATP and citrate
- b) Uses fructose-1-phosphate as substrate
- c) Is the rate-limiting reaction of the glycolytic pathway
- d) Is inhibited by fructose 2, 6-bisphosphate
Explanation:
Phosphofructokinase-1 (PFK-1) is the most important regulatory enzyme in glycolysis. It is considered the committed and rate-limiting step because it is highly regulated and irreversible under physiological conditions.
Why other options are incorrect:
Why other options are incorrect:
- • ATP and Citrate: These act as allosteric inhibitors, not activators. They signal that the cell has high energy levels.
- • Substrate: The substrate for PFK-1 is Fructose-6-phosphate, which is then converted to Fructose-1,6-bisphosphate.
- • Fructose 2,6-bisphosphate: This is actually the most potent activator of PFK-1, not an inhibitor.
Compared to the resting state, vigorously contracting muscle shows:
- a) An increased conversion of pyruvate to lactate
- b) Decreased oxidation of pyruvate to CO2 and water
- c) A decreased NADH/NAD+ ratio
- d) Decreased concentration of AMP
Explanation:
During vigorous contraction, the demand for ATP exceeds the oxygen supply (hypoxia). To maintain glycolysis and continue generating ATP, pyruvate is reduced to lactate by lactate dehydrogenase, which regenerates the NAD+ needed for glycolysis to continue.
Why other options are incorrect:
Why other options are incorrect:
- • Oxidation of Pyruvate: While anaerobic pathways take over, the total flux of aerobic oxidation actually increases to its maximum capacity compared to the resting state.
- • NADH/NAD+ ratio: This ratio actually increases because NADH is being produced rapidly by glycolysis and the TCA cycle, often faster than the electron transport chain can process it.
- • AMP concentration: AMP levels increase significantly during exercise as ATP is consumed and the adenylate kinase reaction ($2 \text{ ADP} \leftrightarrow \text{ATP} + \text{AMP}$) is triggered.
Which one of the following would be expected in pyruvate kinase deficiency?
- a) Increased levels of lactate in the R.B.C
- b) Hemolytic anemia
- c) Decreased ratio of ADP to ATP in R.B.C
- d) Increased phosphorylation of Glucose to Glucose-6-phosphate
Explanation:
Pyruvate kinase deficiency is a common cause of hemolytic anemia. Because Red Blood Cells (RBCs) lack mitochondria, they depend entirely on glycolysis for ATP. A deficiency in pyruvate kinase (the final step of glycolysis) leads to ATP depletion, causing the RBC membrane to lose its integrity and undergo premature destruction in the spleen.
Why other options are incorrect:
Why other options are incorrect:
- • Lactate levels: These would actually be decreased because the glycolytic pathway is blocked before pyruvate (the precursor to lactate) is formed.
- • ADP to ATP ratio: The ratio would increase, not decrease, because ATP production is severely impaired while ADP levels rise.
- • Glucose Phosphorylation: This step (catalyzed by hexokinase) is generally not increased; in fact, the accumulation of upstream intermediates like 2,3-BPG may actually regulate the pathway differently.
Which one of the following statements concerning glucose metabolism is correct?
- a) The conversion of Glucose to lactate occurs only in the R.B.C
- b) Glucose enters most cells by a mechanism in which Na+ and glucose are co-transported
- c) Pyruvate kinase catalyses an irreversible reaction
- d) An elevated level of insulin leads to a decreased level of fructose 2, 6-bisphosphate in hepatocyte
Explanation:
In glycolysis, there are three irreversible steps catalyzed by Hexokinase, Phosphofructokinase-1, and Pyruvate Kinase. These reactions have a large negative Gibbs free energy change ($\Delta G$), making them the primary sites of regulation.
Why other options are incorrect:
Why other options are incorrect:
- • Lactate: While RBCs always produce lactate, other tissues (like skeletal muscle during vigorous exercise) also convert glucose to lactate under anaerobic conditions.
- • Glucose Entry: In most tissues, glucose enters via facilitated diffusion (GLUT transporters). Na+-glucose co-transport (SGLT) occurs primarily in the epithelial cells of the intestine and renal tubules.
- • Insulin & F-2,6-BP: Insulin actually increases the level of fructose 2,6-bisphosphate in hepatocytes by activating PFK-2, which in turn stimulates glycolysis.
Which one of the following compounds cannot give rise to the net synthesis of Glucose?
- a) Lactate
- b) Glycerol
- c) α-ketoglutarate
- d) Acetyl CoA
Explanation:
Acetyl CoA cannot give rise to the net synthesis of glucose in humans because the pyruvate dehydrogenase reaction is irreversible, and the carbons of acetyl CoA are ultimately lost as CO2 in the TCA cycle.
Why other options are incorrect:
Why other options are incorrect:
- • Lactate: Can be converted to pyruvate and then used for gluconeogenesis.
- • Glycerol: Can enter the gluconeogenic pathway through glycerol-3-phosphate and dihydroxyacetone phosphate.
- • α-ketoglutarate: Can be converted to oxaloacetate through the TCA cycle and contribute to glucose synthesis.
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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