Chapter 119 – MCQs for Lab (Biochemistry)

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Chapter 119 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.

5901 to 5950 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 119 (Biochemistry)

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Questions 5901 to 5950

In Ames’ assay, liver homogenate is included in the culture medium because:
  • a) It converts pro-carcinogens into carcinogens
  • b) Liver can metabolise histidine
  • c) Salmonella mainly infects liver
  • d) Liver is very susceptible to cancer
Explanation: The Ames test uses liver homogenate (often called the S9 fraction) to simulate mammalian metabolism. Many chemicals (pro-carcinogens) are not mutagenic by themselves but are converted into active carcinogens by liver enzymes (specifically the cytochrome P450 system).
Why other options are incorrect:
  • • Liver can metabolise histidine: The Ames test uses histidine-dependent Salmonella strains to detect mutations; the liver’s role is chemical activation, not histidine metabolism.
  • • Salmonella mainly infects liver: Salmonella is used as a biological sensor for DNA damage; its infection sites are irrelevant to the chemical mechanism of the assay.
  • • Liver is very susceptible to cancer: While the liver is biologically susceptible to cancer, this is not the functional reason why liver homogenate is added to this specific test.
Bile pigments are present and urobilinogen absent in urine in:
  • a) Haemolytic jaundice
  • b) Hepatocellular jaundice
  • c) Obstructive jaundice
  • d) Crigler-Najjar syndrome
Explanation: In obstructive (post-hepatic) jaundice, the flow of bile into the intestine is blocked. Since urobilinogen is formed in the intestine by bacterial action on bilirubin, its production ceases, resulting in its absence in the urine. Meanwhile, conjugated bilirubin levels rise in the blood and spill into the urine as bile pigments.
Why other options are incorrect:
  • • Haemolytic jaundice: Characterized by increased urobilinogen in urine due to excessive breakdown of red blood cells.
  • • Hepatocellular jaundice: Both bile pigments and urobilinogen are typically present in the urine due to impaired liver function.
  • • Crigler-Najjar syndrome: A disorder of bilirubin conjugation; it does not present with the characteristic bile pigment/urobilinogen profile of obstructive jaundice.
Bile pigments are absent and urobilinogen increased in urine in:
  • a) Haemolytic jaundice
  • b) Hepatocellular jaundice
  • c) Obstructive jaundice
  • d) Rotor’s syndrome
Explanation: In haemolytic jaundice, there is excessive destruction of RBCs, leading to high levels of unconjugated bilirubin. Since unconjugated bilirubin is not water-soluble, it does not appear in urine (hence, no bile pigments). However, it leads to increased production of urobilinogen in the intestine, which is then reabsorbed and excreted in the urine.
Why other options are incorrect:
  • • Hepatocellular jaundice: Typically presents with both bile pigments and urobilinogen in the urine due to damage to liver cells.
  • • Obstructive jaundice: Presents with bile pigments present and urobilinogen absent.
  • • Rotor’s syndrome: A benign hereditary condition characterized by conjugated hyperbilirubinemia, which typically results in the presence of bile pigments in the urine.
In obstructive jaundice, urine shows:
  • a) Absence of bile pigments and urobilinogen
  • b) Presence of bile pigments and urobilinogen
  • c) Absence of bile pigments and presence of urobilinogen
  • d) Presence of bile pigments and absence of urobilinogen
Explanation: In obstructive jaundice, the blockage of the bile duct prevents conjugated bilirubin from reaching the intestine. Therefore, no urobilinogen is formed (leading to its absence in urine). The back-pressure of bile into the blood results in elevated conjugated bilirubin, which is water-soluble and filtered into the urine, appearing as bile pigments.
Why other options are incorrect:
  • • a) & c): These are incorrect because bile pigments (conjugated bilirubin) are always present in obstructive jaundice due to their reflux into the bloodstream.
  • • b): This pattern is more characteristic of hepatocellular jaundice, where both the drainage and the ability to process bilirubin are compromised.
In haemolytic jaundice, urine shows:
  • a) Absence of bile pigments and urobilinogen
  • b) Presence of bile pigments and urobilinogen
  • c) Absence of bile pigments and presence of urobilinogen
  • d) Presence of bile pigments and absence of urobilinogen
Explanation: In haemolytic jaundice, there is an overproduction of unconjugated bilirubin from excessive red blood cell destruction. Since unconjugated bilirubin is bound to albumin and not water-soluble, it cannot be filtered by the kidneys (hence, no bile pigments in urine). However, the liver produces high amounts of conjugated bilirubin, which is excreted into the gut and converted into a large excess of urobilinogen, which is then reabsorbed and excreted in the urine.
Why other options are incorrect:
  • • a) & d): These are incorrect because urobilinogen levels are significantly increased in haemolytic jaundice due to high bilirubin load.
  • • b): This is incorrect because bile pigments (conjugated bilirubin) do not appear in the urine in pure haemolytic cases, as the liver’s conjugation capacity is not initially overwhelmed.
Serum albumin may be decreased in:
  • a) Haemolytic jaundice
  • b) Hepatocellular jaundice
  • c) Obstructive jaundice
  • d) All of these
Explanation: Serum albumin is synthesized exclusively by the liver. In hepatocellular jaundice (such as cirrhosis or severe hepatitis), the liver’s functional capacity is damaged, leading to a decrease in the synthesis of albumin (hypoalbuminemia).
Why other options are incorrect:
  • • Haemolytic jaundice: This is a pre-hepatic condition primarily involving RBC destruction; liver synthetic function is usually preserved.
  • • Obstructive jaundice: This is a post-hepatic condition involving bile duct blockage; the liver’s synthetic function remains intact unless the condition is long-standing and causes secondary liver damage.
  • • All of these: Since albumin synthesis is specific to liver cell health, it is not consistently decreased in haemolytic or uncomplicated obstructive jaundice.
Normal range of serum albumin is:
  • a) 2.0–3.6 gm/dl
  • b) 2.0–3.6 mg/dl
  • c) 3.5–5.5 gm/dl
  • d) 3.5–5.5 mg/dl
Explanation: The normal reference range for serum albumin in a healthy adult is typically 3.5 to 5.5 grams per deciliter (g/dL). Albumin is the most abundant protein in human blood plasma and plays a crucial role in maintaining oncotic pressure and transporting various substances.
Why other options are incorrect:
  • • Units (mg/dl vs gm/dl): Albumin is present in high concentrations, measured in grams per deciliter (g/dL). Measuring in milligrams per deciliter (mg/dL) would represent an incorrect order of magnitude.
  • • Values: The range of 2.0–3.6 g/dL is significantly lower than the standard clinical reference range and usually indicates hypoalbuminemia (malnutrition, liver disease, or kidney loss).
Normal range of serum globulin is:
  • a) 2.0–3.6 mg/dl
  • b) 2.0–3.6 gm/dl
  • c) 3.5–5.5 mg/dl
  • d) 3.5–5.5 gm/dl
Explanation: The normal reference range for serum globulins (the total of alpha, beta, and gamma globulins) is approximately 2.0 to 3.6 g/dL. Globulins are a group of proteins in your blood that play an important role in liver function, blood clotting, and fighting infection.
Why other options are incorrect:
  • • Units (mg/dl vs gm/dl): Like albumin, globulins are measured in grams per deciliter (g/dL); mg/dL values would be misleadingly small.
  • • Values: The range of 3.5–5.5 g/dL corresponds to the normal range for albumin, not globulins.
Serum albumin: globulin ratio is altered in:
  • a) Gilbert’s disease
  • b) Haemolytic jaundice
  • c) Viral hepatitis
  • d) Stones in bile duct
Explanation: In viral hepatitis (a form of hepatocellular damage), the liver’s ability to synthesize albumin is reduced (hypoalbuminemia), while chronic inflammation often leads to an increase in gamma-globulins (hyperglobulinemia). This shift decreases the albumin:globulin (A:G) ratio.
Why other options are incorrect:
  • • Gilbert’s disease: A benign genetic disorder of bilirubin metabolism that does not affect liver protein synthesis.
  • • Haemolytic jaundice: Primarily involves the breakdown of red blood cells; liver synthetic function remains normal.
  • • Stones in bile duct: This is a mechanical obstruction; as long as the liver parenchyma is not significantly damaged, protein synthesis remains normal and the A:G ratio is typically unaffected.
Esterification of cholesterol occurs mainly in:
  • a) Adipose tissue
  • b) Liver
  • c) Muscles
  • d) Kidneys
Explanation: The liver is the primary site for the esterification of cholesterol. This process is catalyzed by the enzyme Lecithin-Cholesterol Acyltransferase (LCAT) in the blood (associated with HDL) and Acyl-CoA:Cholesterol Acyltransferase (ACAT) within the liver cells. Esterification makes cholesterol more hydrophobic, allowing it to be stored within the cell or packaged into lipoproteins.
Why other options are incorrect:
  • • Adipose tissue: Primarily functions in the storage of triglycerides, not the primary site for cholesterol esterification.
  • • Muscles: Utilize cholesterol for cell membrane maintenance but are not the main site for systemic cholesterol processing or esterification.
  • • Kidneys: Primarily involved in filtration and excretion; they do not perform cholesterol esterification as a primary metabolic function.
Galactose intolerance can occur in:
  • a) Haemolytic jaundice
  • b) Hepatocellular jaundice
  • c) Obstructive jaundice
  • d) None of these
Explanation: The liver is the primary site for the metabolism of galactose (converting it into glucose). In hepatocellular jaundice, the functional capacity of the liver cells (hepatocytes) is compromised. This impairment reduces the liver’s ability to process galactose efficiently, leading to galactose intolerance.
Why other options are incorrect:
  • • Haemolytic jaundice: This condition involves the breakdown of red blood cells and does not directly interfere with the liver’s metabolic capacity to handle galactose.
  • • Obstructive jaundice: This is a mechanical obstruction of bile flow. While the bile ducts are blocked, the liver cells themselves are not necessarily dysfunctional in their metabolic pathways (initially), so galactose metabolism remains largely intact.
Prothrombin is synthesised in:
  • a) Erythrocytes
  • b) Reticulo-endothelial cells
  • c) Liver
  • d) Kidneys
Explanation: Prothrombin (Factor II) is a vitamin K-dependent glycoprotein synthesized primarily in the liver. It is a vital component of the blood coagulation cascade, and its production can be impaired in patients with severe liver disease or vitamin K deficiency.
Why other options are incorrect:
  • • Erythrocytes: Red blood cells are involved in gas transport and do not have the protein synthesis machinery to produce clotting factors.
  • • Reticulo-endothelial cells: These are primarily involved in immune defense and the breakdown of old blood cells, not the synthesis of coagulation proteins.
  • • Kidneys: While the kidneys produce hormones like erythropoietin, they are not the site of prothrombin synthesis.
Prothrombin time remains prolonged even after parenterals administration of vitamin K in:
  • a) Haemolytic jaundice
  • b) Liver damage
  • c) Biliary obstruction
  • d) Steatorrhoea
Explanation: Prothrombin synthesis in the liver is dependent on Vitamin K. In liver damage (hepatocellular disease), the hepatocytes are unable to synthesize prothrombin even when sufficient vitamin K is provided. In contrast, conditions like biliary obstruction or steatorrhoea cause vitamin K deficiency due to poor absorption; therefore, parenteral administration usually corrects the prothrombin time.
Why other options are incorrect:
  • • Biliary obstruction & Steatorrhoea: These cause a lack of bile salts needed to absorb fat-soluble Vitamin K. Once Vitamin K is given parenterally (bypassing the gut), synthesis returns to normal.
  • • Haemolytic jaundice: Does not primarily affect the liver’s synthetic capacity for clotting factors.
All the following statements about obstructive jaundice are true except:
  • a) Conjugated bilirubin in serum is normal
  • b) Total bilirubin in serum is raised
  • c) Bile salts are present in urine
  • d) Serum alkaline phosphatase is raised
Explanation: The statement “Conjugated bilirubin in serum is normal” is false (the correct answer to the ‘except’ question). In obstructive jaundice, the flow of bile is blocked, causing conjugated bilirubin to regurgitate into the bloodstream, resulting in significantly elevated levels of serum conjugated (direct) bilirubin.
Why other options are true:
  • • Total bilirubin is raised: This is a hallmark of all types of jaundice.
  • • Bile salts are present in urine: Because of the blockage, bile salts regurgitate into the blood and are filtered into the urine (choluria).
  • • Serum alkaline phosphatase is raised: This enzyme is located in the bile canaliculi; its levels increase significantly due to the back-pressure from the obstruction.
All the following statements about obstructive jaundice are true except:
  • a) Prothrombin time may be prolonged due to impaired absorption of vitamin K
  • b) Serum alkaline phosphatase may be raised due to increased release of the enzyme from liver cells
  • c) Bile salts may enter systemic circulation due to biliary obstruction
  • d) There is no defect in conjugation of bilirubin
Explanation: The statement regarding alkaline phosphatase is technically inaccurate because the elevation is specifically due to increased synthesis and induction of the enzyme in the epithelial cells of the bile ducts in response to cholestasis, followed by regurgitation into the blood—not simply “release from liver cells” (which implies cell damage as seen with AST/ALT).
Why other options are true:
  • • a) Vitamin K is fat-soluble; obstruction prevents bile salts from entering the gut, causing poor fat absorption and Vitamin K deficiency.
  • • c) Obstruction causes “back-pressure,” forcing bile components (including bile salts) into the blood.
  • • d) In obstructive jaundice, the liver parenchyma is functioning; the bilirubin is successfully conjugated, but it cannot be excreted into the gut.
A test to evaluate detoxifying function of liver is:
  • a) Serum albumin: globulin ratio
  • b) Galactose tolerance test
  • c) Hippuric acid test
  • d) Prothrombin time
Explanation: The Hippuric acid test specifically evaluates the detoxifying (synthetic/conjugation) capacity of the liver. In this test, sodium benzoate is administered, which the liver conjugates with glycine to form hippuric acid. The rate at which hippuric acid is excreted in the urine is a measure of the liver’s detoxifying ability.
Why other options are incorrect:
  • • Serum A:G ratio: Evaluates the synthetic (protein) function of the liver.
  • • Galactose tolerance test: Evaluates the metabolic (carbohydrate) function of the liver.
  • • Prothrombin time: Evaluates the synthetic function related to coagulation factors, which is dependent on both liver health and Vitamin K status.
Hippuric acid is formed from:
  • a) Benzoic acid and alanine
  • b) Benzoic acid and glycine
  • c) Glucuronic acid and alanine
  • d) Glucuronic acid and glycine
Explanation: Hippuric acid is the product of the conjugation of benzoic acid with the amino acid glycine in the liver. This is a classic example of phase II detoxification, where the liver renders a toxic or foreign substance (like benzoate) more water-soluble for excretion in the urine.
Why other options are incorrect:
  • • Alanine: Is not involved in the conjugation reaction that forms hippuric acid.
  • • Glucuronic acid: While glucuronic acid is involved in other important conjugation reactions in the liver (such as with bilirubin), it is not the substrate used to form hippuric acid.
An enzyme which is excreted in urine is:
  • a) Lactate dehydrogenase
  • b) Amylase
  • c) Ornithine transcarbamoylase
  • d) None of these
Explanation: Amylase has a relatively low molecular weight, which allows it to be filtered by the renal glomeruli and excreted in the urine. Measuring urinary amylase is a common clinical diagnostic procedure, particularly when assessing cases of acute pancreatitis, as it often remains elevated in the urine longer than in the blood.
Why other options are incorrect:
  • • Lactate dehydrogenase: A large intracellular enzyme; its presence in urine usually indicates damage to renal cells rather than normal excretion.
  • • Ornithine transcarbamoylase: A mitochondrial enzyme involved in the urea cycle; it is not typically found in the urine in healthy individuals.
Serum gamma glutamyl transpeptidase (GGT) is raised in:
  • a) Haemolytic jaundice
  • b) Myocardial infarction
  • c) Alcoholic hepatitis
  • d) Acute cholecystitis
Explanation: Serum Gamma-Glutamyl Transpeptidase (GGT) is a very sensitive marker for liver disease. It is particularly elevated in alcoholic liver disease (like alcoholic hepatitis) because alcohol induces the synthesis of this enzyme in the liver. It is also a sensitive marker for hepatobiliary obstruction.
Why other options are incorrect:
  • • Haemolytic jaundice: Primarily involves the breakdown of red blood cells and does not cause induction of GGT.
  • • Myocardial infarction: Involves heart muscle damage; cardiac enzymes like Troponin or CK-MB are raised, not GGT.
  • • Acute cholecystitis: While it can cause biliary obstruction (which may raise GGT), GGT is specifically considered the “gold standard” marker for chronic alcohol consumption and alcoholic liver damage.
Urea clearance is the:
  • a) Amount of urea excreted per minute
  • b) Amount of urea present in 100 ml of urine
  • c) Volume of blood cleared of urea in one minute
  • d) Amount of urea filtered by glomeruli in one minute
Explanation: Renal clearance is defined as the volume of plasma from which a substance is completely removed (cleared) by the kidneys per unit of time. Therefore, urea clearance is the theoretical volume of blood (plasma) that is completely cleared of urea by the kidneys in one minute, typically expressed in ml/min.
Why other options are incorrect:
  • • a) This describes the “excretion rate,” not the clearance.
  • • b) This is simply the concentration of urea in the urine.
  • • d) This describes the “filtered load” of urea, which is higher than the clearance because a significant portion of filtered urea is reabsorbed back into the blood by the renal tubules.
Inulin clearance is a measure of:
  • a) Glomerular filtration rate
  • b) Tubular secretion flow
  • c) Tubular reabsorption rate
  • d) Renal plasma flow
Explanation: Inulin is an ideal substance for measuring the Glomerular Filtration Rate (GFR) because it is freely filtered by the glomeruli and is neither reabsorbed, secreted, nor metabolized by the renal tubules. Therefore, the amount of inulin cleared from the plasma is exactly equal to the amount filtered by the glomeruli.
Why other options are incorrect:
  • • Tubular secretion/reabsorption: Inulin is not involved in either process, making it a “gold standard” marker for GFR specifically.
  • • Renal plasma flow: This is typically measured using substances like Para-aminohippuric acid (PAH), which is both filtered and actively secreted by the tubules.
Phenolsulphonephthalein (PSP) excretion test is an indicator of:
  • a) Glomerular filtration
  • b) Tubular secretion
  • c) Tubular reabsorption
  • d) Renal blood flow
Explanation: The Phenolsulphonephthalein (PSP) excretion test is a classic clinical test used to evaluate the secretory capacity of the renal tubules. Once injected, PSP is rapidly excreted by the kidneys, primarily through active transport by the proximal convoluted tubules. A decrease in the rate of PSP excretion is a sensitive indicator of impaired tubular function.
Why other options are incorrect:
  • • Glomerular filtration: While PSP is also filtered, its rapid rate of clearance is largely due to active tubular secretion, making it a test of tubular function rather than GFR.
  • • Tubular reabsorption: PSP is not significantly reabsorbed by the tubules.
  • • Renal blood flow: While blood flow affects the delivery of substances to the kidneys, other substances like Para-aminohippuric acid (PAH) are more specific markers for total renal plasma flow.
Para-amino hippurate (PAH) excretion test is an indicator of:
  • a) Glomerular filtration
  • b) Tubular secretion
  • c) Tubular reabsorption
  • d) Renal plasma flow
Explanation: Para-amino hippurate (PAH) is almost completely cleared from the renal plasma in a single pass through the kidneys, both by glomerular filtration and, more importantly, by extensive active tubular secretion. Because it is almost entirely removed from the blood that perfuses the functioning renal tissue, the clearance of PAH is used to estimate the Effective Renal Plasma Flow (ERPF).
Why other options are incorrect:
  • • Glomerular filtration: While PAH is filtered, its total clearance is determined by the plasma flow, not just the filtration rate.
  • • Tubular secretion: While the mechanism of PAH clearance involves tubular secretion, the clinical purpose of the test is to measure renal plasma flow.
  • • Tubular reabsorption: PAH is not reabsorbed by the tubules.
Renal plasma flow (RPF) of an average adult man is:
  • a) 120–130 ml/minute
  • b) 325–350 ml/minute
  • c) 480–520 ml/minute
  • d) 560–830 ml/minute
Explanation: The normal Renal Plasma Flow (RPF) for an average adult male is approximately 600–700 ml/min, which falls within the range of 560–830 ml/min. This represents the volume of plasma delivered to the kidneys per minute, which is critical for maintaining glomerular filtration and renal function.
Why other options are incorrect:
  • • 120–130 ml/minute: This range corresponds to the normal Glomerular Filtration Rate (GFR) for an adult male, not the plasma flow.
  • • 325–350 ml/minute: This is significantly lower than the actual physiological plasma flow to the kidneys.
  • • 480–520 ml/minute: While closer to the lower end, the standard physiological range for RPF is generally cited as slightly higher (averaging ~600-650 ml/min).
Filtration fraction can be calculated from:
  • a) Standard urea clearance and PSP excretion
  • b) Maximum urea clearance and PSP excretion
  • c) Maximum urea clearance and PAH clearance
  • d) Inulin clearance and PAH clearance
Explanation: The Filtration Fraction (FF) is the ratio of the Glomerular Filtration Rate (GFR) to the Renal Plasma Flow (RPF). Mathematically: FF = GFR / RPF. Since Inulin clearance accurately measures GFR and PAH clearance accurately measures RPF, the filtration fraction is calculated by dividing Inulin clearance by PAH clearance.
Why other options are incorrect:
  • • Urea clearance: It is not as accurate as Inulin clearance because a significant amount of urea is reabsorbed by the tubules.
  • • PSP excretion: This is a test of tubular secretory function, not a direct quantitative measure of GFR or RPF required for the filtration fraction calculation.
Normal filtration fraction is about:
  • a) 0.2
  • b) 0.4
  • c) 0.6
  • d) 0.8
Explanation: The filtration fraction is the ratio of Glomerular Filtration Rate (GFR) to Renal Plasma Flow (RPF). With a normal GFR of approximately 125 ml/min and a normal RPF of approximately 625 ml/min, the calculation 125/625 yields a value of 0.2 (or 20%). This means about 20% of the plasma entering the kidneys is filtered into the Bowman’s capsule.
Why other options are incorrect:
  • • 0.4, 0.6, 0.8: These values represent significantly higher percentages of filtration than the normal physiological capacity of the glomeruli, which would indicate abnormal kidney function or hemodynamics.
Filtration fraction is increased in:
  • a) Acute glomerulonephritis
  • b) Chronic glomerulonephritis
  • c) Hypertension
  • d) Hypotension
Explanation: Filtration fraction (FF = GFR/RPF) increases in conditions like hypertension (or states with increased Angiotensin II). Angiotensin II preferentially constricts the efferent arterioles of the glomerulus. This increases the glomerular capillary hydrostatic pressure (maintaining GFR) while reducing the blood flow out of the glomerulus, thereby decreasing RPF. When GFR remains stable while RPF decreases, the filtration fraction rises.
Why other options are incorrect:
  • • Glomerulonephritis (Acute/Chronic): These conditions damage the filtration membrane, which typically causes a decrease in GFR, leading to a decreased or variable filtration fraction.
  • • Hypotension: Typically leads to a decrease in GFR because the drop in systemic pressure is so significant that it cannot be fully compensated for by arteriolar adjustments, resulting in a lower filtration rate relative to plasma flow.
  1. Filtration fraction is increased in
    • Acute glomerulonephritis
    • Chronic glomerulonephritis
    • Hypertension✔
    • Hypotension
Esters of fatty acids with higher alcohols other than glycerol are said to be:
  • a) Waxes
  • b) Fats
  • c) Both (Waxes) and (Fats)
  • d) None of these
Explanation: Waxes are defined as esters formed from long-chain fatty acids and high-molecular-weight monohydric alcohols (alcohols other than glycerol). Unlike fats (which are esters of fatty acids with the trihydric alcohol, glycerol), waxes are typically solid at room temperature and serve as protective coatings in plants and animals.
Why other options are incorrect:
  • • Fats: These are specifically esters of fatty acids with glycerol (triglycerides).
  • • Both (Waxes) and (Fats): This is incorrect because fats must contain glycerol, whereas waxes contain other types of alcohols.
The combination of an amino alcohol, fatty acid and sialic acid form:
  • a) Phospholipids
  • b) Sulpholipids
  • c) Glycolipids
  • d) Aminolipids
Explanation: The combination of an amino alcohol (like sphingosine), a fatty acid, and a carbohydrate group containing sialic acid (N-acetylneuraminic acid) forms Gangliosides, which are a complex class of glycolipids (specifically glycosphingolipids).
Why other options are incorrect:
  • • Phospholipids: These require a phosphate group in their structure.
  • • Sulpholipids: These contain a sulfate group.
  • • Aminolipids: This is a broad term for lipids containing nitrogenous bases, but it does not specifically define the complex carbohydrate-containing structure described here.
Hydrolysis of fats by alkali is called:
  • a) Saponification number
  • b) Saponification
  • c) Both (Saponification number) and (Saponification)
  • d) None of these
Explanation: The process of hydrolyzing fats (triglycerides) using a strong alkali (such as NaOH or KOH) to produce glycerol and the salts of fatty acids (soaps) is known as Saponification.
Why other options are incorrect:
  • • Saponification number: This is a quantitative measure (the number of milligrams of KOH required to saponify one gram of fat), not the name of the chemical process itself.
  • • Both: As the definition distinguishes between the process and the analytical measure, “both” is technically incorrect in this context.
The number of milliliters of 0.1 N KOH required to neutralize the insoluble fatty acids from 5 gms of fat is called:
  • a) Acid number
  • b) Acetyl number
  • c) Halogenation
  • d) Polenske number
Explanation: The Polenske number is a measure of the volatile, water-insoluble fatty acids (such as caprylic, capric, and lauric acids) present in a fat. It is defined as the number of milliliters of 0.1 N KOH required to neutralize the insoluble fatty acids obtained from 5 grams of fat.
Why other options are incorrect:
  • • Acid number: The number of milligrams of KOH required to neutralize the free fatty acids in 1 gram of fat.
  • • Acetyl number: A measure of the number of hydroxyl groups present in a fat.
  • • Halogenation: This refers to the addition of halogens across double bonds, typically used to determine the iodine number (degree of unsaturation).
The rate of fatty acid oxidation is increased by:
  • a) Phospholipids
  • b) Glycolipids
  • c) Aminolipids
  • d) All of these
Explanation: The rate of fatty acid oxidation can be influenced by various membrane lipids, including phospholipids, glycolipids, and aminolipids. These lipids participate in cellular membrane structure and metabolism, and changes in lipid metabolism can affect fatty acid oxidation. Therefore, the answer is all of these.
Why other options are incorrect:
  • • Phospholipids: Participate in membrane structure and lipid metabolism but are not the only lipid class that can influence fatty acid oxidation.
  • • Glycolipids: Are involved in membrane functions and lipid metabolism but are not the sole factor affecting fatty acid oxidation.
  • • Aminolipids: Also participate in cellular lipid metabolism, but considering the options, all three lipid classes are included.
Lecithin contains a nitrogenous base named as
  • a) Ethanolamine
  • b) Choline
  • c) Inositol
  • d) All of these
Explanation: Lecithin is chemically known as Phosphatidylcholine. It is a phospholipid that contains glycerol, two fatty acids, a phosphate group, and the nitrogenous base Choline.
Why other options are incorrect:
  • • Ethanolamine: Found in Phosphatidylethanolamine (Cephalin), not Lecithin.
  • • Inositol: Found in Phosphatidylinositol; it is a cyclic sugar alcohol and not a nitrogenous base.
  • • All of these: Incorrect, as only Choline is the nitrogenous base present in Lecithin.
Lecithins contain an unsaturated fatty acid at position:
  • a) α
  • b) α and β
  • c) β
  • d) None of these
Explanation: In the structure of lecithin (phosphatidylcholine), the fatty acid at the β-position (carbon-2 of the glycerol backbone) is typically an unsaturated fatty acid, while the fatty acid at the α-position (carbon-1) is usually saturated.
Why other options are incorrect:
  • • α (Alpha): The α-position in lecithin is generally occupied by a saturated fatty acid.
  • • α and β: While both positions hold fatty acids, the specific structural characteristic of lecithin is the presence of the unsaturated fatty acid specifically at the β-carbon.
Lecithins are soluble in ordinary solvents except:
  • a) Benzene
  • b) Ethyl alcohol
  • c) Methyl alcohol
  • d) Acetone
Explanation: Lecithins (phosphatidylcholines) are soluble in most organic solvents like benzene, ether, chloroform, and alcohol. However, they are characteristically insoluble in acetone. This specific property is often used in laboratories to precipitate lecithin out of lipid mixtures.
Why other options are incorrect:
  • • Benzene: A non-polar organic solvent in which lecithin dissolves readily.
  • • Ethyl alcohol & Methyl alcohol: Lecithin is soluble in these polar organic solvents.
  1. Lecithins are soluble in ordinary solvents except
    • Benzene
    • Ethyl alcohol
    • Methyl alcohol
    • Acetone ✔
Lecithins combine with protein to form:
  • a) Phosphoprotein
  • b) Mucoprotein
  • c) Lipoprotein
  • d) Glycoprotein
Explanation: Lecithin is a lipid (specifically a phospholipid). When lipids combine with proteins, the resulting conjugated protein is known as a Lipoprotein. Lipoproteins are essential for the transport of lipids in the blood and the structural integrity of cell membranes.
Why other options are incorrect:
  • • Phosphoprotein: Proteins that contain a phosphate group attached to amino acid side chains, not specifically a lipid-protein complex.
  • • Mucoprotein: A complex of protein and polysaccharides (mucopolysaccharides).
  • • Glycoprotein: A protein with carbohydrate groups covalently attached.
Instead of ester link plasmalogens possess an other link in position:
  • a) α
  • b) β
  • c) γ
  • d) None of these
Explanation: Plasmalogens are a unique class of phospholipids. Unlike typical phospholipids that have ester linkages at both the α and β positions of glycerol, plasmalogens possess an ether linkage (specifically a vinyl ether bond) at the α-position (carbon-1).
Why other options are incorrect:
  • • β (Beta): The β-position in plasmalogens typically contains a standard fatty acid via an ester linkage.
  • • γ (Gamma): Glycerol does not have a “gamma” position; carbon-3 is usually occupied by the phosphate group.
The alkyl radical in plasmalogen is an alcohol:
  • a) Saturated
  • b) Unsaturated
  • c) Both Saturated and Unsaturated
  • d) None of these
Explanation: Plasmalogens are phospholipids containing a vinyl ether linkage at the sn-1 position. The alkyl chain is derived from an unsaturated alcohol, forming the characteristic ether bond of plasmalogens.
Why other options are incorrect:
  • • Saturated: A saturated alcohol would not provide the characteristic double bond of the vinyl ether linkage in plasmalogens.
  • • Both Saturated and Unsaturated: The characteristic alkyl radical of plasmalogens is associated with an unsaturated alcohol.
  • • None of these: Incorrect because the alkyl radical is derived from an unsaturated alcohol.
The concentration of sphingomyelins are increased in:
  • a) Gaucher’s disease
  • b) Fabry’s disease
  • c) Fabrile disease
  • d) Niemann-Pick disease
Explanation: Niemann-Pick disease is a lysosomal storage disorder caused by a deficiency of the enzyme sphingomyelinase. This deficiency leads to the accumulation of sphingomyelin in various tissues, particularly in the liver, spleen, and brain.
Why other options are incorrect:
  • • Gaucher’s disease: Characterized by the accumulation of glucocerebroside due to a deficiency of glucocerebrosidase.
  • • Fabry’s disease: Characterized by the accumulation of globotriaosylceramide (ceramide trihexoside) due to a deficiency of alpha-galactosidase A.
  • • Fabrile disease: This is not a recognized sphingolipid storage disease (likely a distractor).
Sphingomyelins contain a complex amino alcohol named as:
  • a) Serine
  • b) Lysolecithin
  • c) Sphingosine
  • d) GlycoL
Explanation: Sphingomyelins are a type of sphingolipid. Their structural backbone is Sphingosine, a long-chain amino alcohol. This is combined with a fatty acid to form ceramide, which is then linked to a phosphocholine group to complete the sphingomyelin molecule.
Why other options are incorrect:
  • • Serine: While it is a precursor used in the biosynthesis of sphingosine, it is not the structural amino alcohol backbone found in the final sphingomyelin molecule.
  • • Lysolecithin: A derivative of lecithin (phosphatidylcholine) formed by the removal of a fatty acid; not an amino alcohol.
  • • GlycoL: Not a standard biochemical term for an amino alcohol.
The types of sphingomyelins are:
  • a) 1
  • b) 3
  • c) 4
  • d) 5
Explanation: Sphingomyelins are commonly classified into three types based on the fatty acid component attached to sphingosine. These forms differ in their fatty acid composition and distribution in tissues.
Why other options are incorrect:
  • • 1: There is more than one type of sphingomyelin.
  • • 4: The commonly described classification recognizes three types.
  • • 5: Five is not the standard number of types in this classification.
Glycolipids contain an amino alcohol:
  • a) Sphingosine
  • b) Iso-sphingosine
  • c) Both (Sphingosine) and (Iso-sphingosine)
  • d) None of these
Explanation: Glycolipids, particularly glycosphingolipids, contain the amino alcohol sphingosine as part of their ceramide backbone. Some sphingolipids may contain related long-chain bases such as dihydrosphingosine, but the principal amino alcohol in glycolipids is sphingosine.
Why other options are incorrect:
  • • Sphingosine: This is the principal amino alcohol component of glycosphingolipids.
  • • Iso-sphingosine: It is not the usual amino alcohol component of the common glycolipids described in biochemistry.
  • • Both: Incorrect because the standard structure of glycolipids contains sphingosine rather than both sphingosine and iso-sphingosine.
Cerebrosides may also be classified as:
  • a) Sphingolipids
  • b) Sulpholipids
  • c) Aminolipids
  • d) Glycolipids
Explanation: Cerebrosides are glycolipids composed of a ceramide (sphingosine + fatty acid) linked to a single sugar residue, usually glucose or galactose. Therefore, they are classified as glycolipids.
Why other options are incorrect:
  • • Sphingolipids: Cerebrosides are structurally sphingolipids, but the classification specifically asked for here is glycolipids.
  • • Sulpholipids: These contain sulfate groups and are distinct from cerebrosides.
  • • Aminolipids: Cerebrosides are not classified as aminolipids.
Gaucher’s disease is characterized specially by the increase in:
  • a) Lignoceric acid
  • b) Nervonic acid
  • c) Cerebomic acid
  • d) Hydroxynervonic acid
Explanation: Gaucher’s disease is a lysosomal storage disorder caused by deficiency of β-glucocerebrosidase. It leads to accumulation of glucocerebroside (glucosylceramide), particularly in macrophages. Among the given fatty acids, lignoceric acid is the characteristic increase associated with Gaucher’s disease.
Why other options are incorrect:
  • • Nervonic acid: Not the characteristic fatty acid increased in Gaucher’s disease.
  • • Cerebomic acid: Not specifically associated with Gaucher’s disease.
  • • Hydroxynervonic acid: Not the characteristic increase seen in Gaucher’s disease.
Gangliosides are the glycolipids occurring in:
  • a) Brain
  • b) Liver
  • c) Kidney
  • d) Muscle
Explanation: Gangliosides are complex glycolipids that are especially abundant in the membranes of nerve cells and are therefore found predominantly in the brain and nervous tissue.
Why other options are incorrect:
  • • Liver: Contains glycolipids but is not the major site associated with gangliosides.
  • • Kidney: Contains membrane glycolipids, but gangliosides are particularly concentrated in nervous tissue.
  • • Muscle: Contains membrane lipids but is not a major site of ganglioside accumulation.
Lipoprotein present in cell membrane is by nature:
  • a) Hydrophilic
  • b) Hydrophobic
  • c) Both (Hydrophilic) and (Hydrophobic)
  • d) None of these
Explanation: Membrane lipoproteins contain both hydrophilic and hydrophobic regions. Their hydrophobic portions interact with the lipid bilayer, while hydrophilic portions interact with the aqueous environment. Therefore, they are amphipathic in nature.
Why other options are incorrect:
  • • Hydrophilic: Only describes the water-attracting portion; membrane lipoproteins also contain hydrophobic regions.
  • • Hydrophobic: Only describes the lipid-associated portion; membrane lipoproteins also have hydrophilic regions.
  • • None of these: Incorrect because membrane lipoproteins have both hydrophilic and hydrophobic characteristics.
The density of lipoproteins increases as the protein content:
  • a) Increases
  • b) Decreases
  • c) Highly decreases
  • d) Slightly and promptly decreases
Explanation: The density of lipoproteins increases as their protein content increases. Protein is denser than lipid, so lipoproteins with a higher proportion of protein have greater density.
Why other options are incorrect:
  • • Decreases: A decrease in protein content generally results in lower lipoprotein density.
  • • Highly decreases: A marked decrease in protein content would lower, rather than increase, lipoprotein density.
  • • Slightly and promptly decreases: Decreasing protein content does not cause an increase in lipoprotein density.
Lipoproteins may be identified more accurately by means of:
  • a) Electrophoresis
  • b) Ultra centrifugation
  • c) Centrifugation
  • d) Immunoelectrophoresis
Explanation: Lipoproteins are identified and separated more accurately by ultracentrifugation based on their differences in density. Different lipoprotein classes such as chylomicrons, VLDL, LDL, and HDL have characteristic densities.
Why other options are incorrect:
  • • Electrophoresis: Can separate lipoproteins based on their electrophoretic mobility but is less accurate for identification than ultracentrifugation.
  • • Centrifugation: Ordinary centrifugation does not provide the precise density-based separation required for accurate lipoprotein identification.
  • • Immunoelectrophoresis: Primarily identifies proteins using antigen-antibody reactions and is not the standard method for classifying lipoproteins by density.
Very low density lipoproteins are also known as:
  • a) β-lipoproteins
  • b) Pre β-lipoproteins
  • c) α-lipoproteins
  • d) None of these
Explanation: In electrophoresis, Very Low Density Lipoproteins (VLDL) migrate to the position between the alpha and beta bands, which is why they are specifically classified as Pre-β-lipoproteins.
Why other options are incorrect:
  • • β-lipoproteins: This term refers to Low Density Lipoproteins (LDL).
  • • α-lipoproteins: This term refers to High Density Lipoproteins (HDL).
  • • None of these: Incorrect, as VLDL is synonymous with the pre-beta migration pattern.
The protein moiety of lipoprotein is known as:
  • a) Apoprotein
  • b) Pre-protein
  • c) Post-protein
  • d) Pseudoprotein
Explanation: The protein component of a lipoprotein is called an Apoprotein (or apolipoprotein). These proteins serve as structural components, enzyme cofactors, and ligands for cell-surface receptors.
Why other options are incorrect:
  • • Pre-protein: Usually refers to a precursor protein containing a signal peptide that is cleaved during maturation.
  • • Post-protein: This is not a standard biological term used to describe lipoprotein components.
  • • Pseudoprotein: This term is not used in biochemistry to describe the constituents of lipoproteins.

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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