Chemistry Quiz 3 Nov 14-2012
(127).webp "Chemistry Quiz 3 Nov 14-2012 - Quiz")
- 1.
Which of the following statements is false regarding competitive enzyme inhibitors?
- A.
The inhibitor has a similar molecular structure as the natural substrate
- B.
The inhibitor affects the Km
- C.
The inhibitor affect is overcome at high concentrations of normal substrate
- D.
The inhibo=itor affects Vmax
Correct Answer
D. The inhibo=itor affects VmaxExplanation
The false statement regarding competitive enzyme inhibitors is that the inhibitor affects Vmax. In competitive inhibition, the inhibitor competes with the natural substrate for the active site of the enzyme. This competition can be overcome by increasing the concentration of the natural substrate, which allows the enzyme to reach its maximum velocity (Vmax). However, the inhibitor does not directly affect the maximum velocity of the enzyme-catalyzed reaction.Rate this question:
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- 2.
Vmax can be affected withouth alterinf Km in which of the following situations?
- A.
Increase the amount of enzyme
- B.
Presence of an irreversible inhibitor
- C.
Substrate concentrations is in excess
- D.
Presence of a competitive inhibitor
Correct Answer
A. Increase the amount of enzymeExplanation
Increasing the amount of enzyme can affect Vmax without altering Km because Vmax is determined by the maximum rate at which the enzyme can convert substrate to product, while Km is a measure of the affinity of the enzyme for the substrate. By increasing the amount of enzyme, there are more active sites available for the substrate to bind to and be converted to product, resulting in an increase in Vmax. However, the affinity of the enzyme for the substrate, represented by Km, remains unchanged.Rate this question:
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- 3.
To assure that zero order kinetics s maintained in an enzyme reaction, the substrate concentration should be:
- A.
Equal to the Km
- B.
Less than the Km
- C.
At least 10 times greater than the Km
- D.
Equal to 1/Km
Correct Answer
C. At least 10 times greater than the KmExplanation
To maintain zero order kinetics in an enzyme reaction, the substrate concentration should be at least 10 times greater than the Km. Zero order kinetics refers to a reaction rate that is independent of substrate concentration. By ensuring that the substrate concentration is significantly higher than the Km, the enzyme is saturated with substrate molecules, allowing the reaction to proceed at a constant rate. If the substrate concentration is equal to or less than the Km, the reaction will follow first-order kinetics, where the rate is dependent on substrate concentration. Similarly, if the substrate concentration is equal to 1/Km, the reaction will follow second-order kinetics.Rate this question:
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- 4.
The point at which an enzyme reaction is proceeding at the greatest rate is:
- A.
The Michaelis constant (Km)
- B.
Zero order kinetics
- C.
First order kinetics
- D.
Point where the rate of the reaction is dependent on the substrate concentration
Correct Answer
B. Zero order kineticsExplanation
Zero order kinetics refers to a reaction where the rate of the reaction is independent of the substrate concentration. In this case, the reaction is proceeding at the greatest rate, meaning that the rate of the reaction is constant regardless of the substrate concentration. This is in contrast to first order kinetics, where the rate of the reaction is directly proportional to the substrate concentration, and the Michaelis constant (Km), which represents the substrate concentration at which the reaction rate is half of its maximum value. Therefore, zero order kinetics is the correct answer in this case.Rate this question:
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- 5.
Isoenzymes are:
- A.
Multiple molecular forms of an enzye family that catalze the same reaction
- B.
Different enzymes which exhibit the same enzymatic specificity
- C.
Multiple molecular forms of different enzymes which catalyze the same reaction
- D.
Different enzymes which exhibit the same electrophoretic mobility
Correct Answer
A. Multiple molecular forms of an enzye family that catalze the same reactionExplanation
Isoenzymes are multiple molecular forms of an enzyme family that catalyze the same reaction. This means that different forms of the same enzyme can exist, each with slightly different molecular structures, but they all perform the same function of catalyzing a specific reaction. These isoenzymes may have different properties, such as different pH optima or different kinetic parameters, but they ultimately carry out the same biochemical reaction.Rate this question:
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- 6.
Ethanol treatment in methanol intoxication is an example of?
- A.
Non-competitive inhibition
- B.
Competitive inhibition
- C.
Irreversible inhibition
- D.
Enzyme induction
Correct Answer
B. Competitive inhibitionExplanation
Ethanol treatment in methanol intoxication is an example of competitive inhibition. Competitive inhibition occurs when a substance, in this case ethanol, competes with the substrate, methanol, for the active site of the enzyme. By binding to the active site, ethanol prevents methanol from binding and being metabolized by the enzyme, effectively reducing its toxic effects. This type of inhibition can be overcome by increasing the concentration of the substrate, as it will outcompete the inhibitor for the active site.Rate this question:
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- 7.
A non-competitive inhibitor:
- A.
Increases Vmax and maintains Km the same
- B.
Does not affect Vmax or Km
- C.
Decreases Vmax and decreases Km
- D.
Decreases Vmax and maintains Km
Correct Answer
D. Decreases Vmax and maintains KmExplanation
A non-competitive inhibitor decreases Vmax because it binds to the enzyme at a site other than the active site, causing a conformational change that reduces the enzyme's catalytic activity. However, it does not affect Km, which is the substrate concentration at which the enzyme achieves half of its maximum velocity. This is because the inhibitor does not compete with the substrate for binding at the active site, so the affinity of the enzyme for the substrate remains the same.Rate this question:
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- 8.
In calcium methods, specimens are acidified:
- A.
To set pH for colour reaction
- B.
To free bound calcium from protein
- C.
To deproteinize specimen
- D.
To analyze organic calcium only
Correct Answer
B. To free bound calcium from proteinExplanation
In calcium methods, specimens are acidified to free bound calcium from protein. This is because calcium ions can bind to proteins in the specimen, making it difficult to accurately measure the concentration of calcium. By acidifying the specimen, the bound calcium is released from the protein, allowing for a more accurate analysis of the calcium levels.Rate this question:
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- 9.
A patient had a blood specimen collected in a heparin-containing tube for calcium and magnesium determination. Upon centrifugation, the serum appeared hemolyzed. How would this affect magnesium?
- A.
There would be no effect on the magnesium value
- B.
The magnesium would form complexes with the hgb end the results would indicate a false negative value
- C.
The magnesium is incorporated into the bilirubin molecule, leading to a false negative value
- D.
Because erythrocytes contain magnesium, hemolysis would increase its apparent value
Correct Answer
D. Because erythrocytes contain magnesium, hemolysis would increase its apparent value -
- 10.
In measuring iron, which of the following statement is most correct?
- A.
The specimen is acidified in order to release iron from transferrin
- B.
The specimen if acidified to oxidize Fe(3+)to Fe(2+)
- C.
The specimen is acidified to release iron from transferrin and to oxidize Fe(3+) to Fe(2+)
- D.
The specimen is acidified ot release iron from transferrin and to reduce Fe(3+) to Fe(2+)
Correct Answer
D. The specimen is acidified ot release iron from transferrin and to reduce Fe(3+) to Fe(2+)Explanation
The correct answer is "the specimen is acidified to release iron from transferrin and to reduce Fe(3+) to Fe(2+)." Acidification of the specimen serves two purposes: it helps to release iron from transferrin, which is a protein that binds to iron, and it also helps to reduce Fe(3+) to Fe(2+). Fe(3+) is the oxidized form of iron, while Fe(2+) is the reduced form. By reducing Fe(3+) to Fe(2+), it allows for more accurate measurement of iron levels in the specimen.Rate this question:
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- 11.
Which of the following reagents is used to measure phosphorus?
- A.
Cresolphtalein complexone
- B.
Ammonium molybdate
- C.
Ferrozine
- D.
Arsenoazo II
Correct Answer
B. Ammonium molybdateExplanation
Ammonium molybdate is used to measure phosphorus. It is commonly used in the molybdenum blue method, where it reacts with phosphorus to form a blue complex. This complex is then measured spectrophotometrically to determine the concentration of phosphorus in a sample. Ammonium molybdate is a reliable and sensitive reagent for phosphorus measurement, making it a popular choice in various analytical techniques.Rate this question:
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- 12.
Kinase enzymes while carboxylase enzymes
- A.
Transfer a phosphate, remove a carboxyl group
- B.
Remove hydrogen, add a carboxyl group
- C.
Remove hydrogen, remove carboxyl group
- D.
Remove phosphate, add carboxyl
Correct Answer
A. Transfer a phosphate, remove a carboxyl groupExplanation
Kinase enzymes transfer a phosphate group from one molecule to another. This process is known as phosphorylation. On the other hand, carboxylase enzymes remove a carboxyl group from a molecule. Therefore, the correct answer "Transfer a phosphate, remove a carboxyl group" accurately describes the functions of kinase and carboxylase enzymes.Rate this question:
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- 13.
The enzyme catalyzed reaction (below) requires what type of enzyme? A-X + B --> A + B-X
- A.
Hydrolase
- B.
Lyase
- C.
Isomerase
- D.
Transferase
Correct Answer
D. TransferaseExplanation
The reaction shown involves the transfer of a functional group (X) from molecule A to molecule B, resulting in the formation of A-X and B-X. This type of reaction is characteristic of transferase enzymes, which facilitate the transfer of functional groups between molecules. Therefore, the correct answer is Transferase.Rate this question:
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- 14.
A DNA ligase enzyme joins group of a nucleotide in a DNA strand with group of an adjacent nucleotide to from bond
- A.
Hydroxyl, phosphoryl, phospho-ester
- B.
Carboxyl, amino, peptide
- C.
Hydroxyl, hydroxyl, glycosidic
- D.
Hydroxy, amino, amide
Correct Answer
A. Hydroxyl, phosphoryl, phospho-esterExplanation
A DNA ligase enzyme joins the hydroxyl group of a nucleotide in a DNA strand with the phosphoryl group of an adjacent nucleotide to form a phospho-ester bond.Rate this question:
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- 15.
In the Jendrassik-Grof method for the determination of total bilirubin, the colour reaction involves which of the following reactants?
- A.
Alkaline tartare
- B.
Caffeine sodium benzoate
- C.
Diazosulphanilic acid
- D.
Ascorbic acid
Correct Answer
C. Diazosulphanilic acidExplanation
The Jendrassik-Grof method for the determination of total bilirubin involves a color reaction. Diazosulphanilic acid is one of the reactants used in this method. Therefore, it is the correct answer.Rate this question:
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- 16.
Under specified conditions of pH and temperature, the international unit (IU) is defined as:
- A.
The enzyme activity that ccatalyzes the conversion of 1umol of substrate in 1 min under standard conditions
- B.
The enzyme activity that ccatalyzes the conversion of 1mol of substrate in 1 min under standard conditions
- C.
The enzyme activity that ccatalyzes the conversion of 1umol of product in 1 min under standard conditions
- D.
The enzyme activity that ccatalyzes the conversion of 1umol of product in 1 min under standard conditions
Correct Answer
A. The enzyme activity that ccatalyzes the conversion of 1umol of substrate in 1 min under standard conditionsExplanation
The correct answer is "The enzyme activity that catalyzes the conversion of 1umol of substrate in 1 min under standard conditions." This answer is correct because the international unit (IU) is a measure of enzyme activity, specifically the amount of enzyme that catalyzes a certain reaction. In this case, the IU is defined as the amount of enzyme that converts 1umol (micromole) of substrate in 1 minute under standard conditions of pH and temperature. This definition allows for standardized measurement and comparison of enzyme activity across different experiments and laboratories.Rate this question:
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- 17.
Which anticoagulant is the best anticoagulant for serum glucose analysis because it inhibits glycolysis?
- A.
Sodium oxalate
- B.
EDTA
- C.
Sodium fluoride
- D.
Heparin
Correct Answer
C. Sodium fluorideExplanation
Sodium fluoride is the best anticoagulant for serum glucose analysis because it inhibits glycolysis. Glycolysis is the breakdown of glucose, and by inhibiting this process, sodium fluoride helps to preserve the glucose levels in the serum sample. This allows for more accurate analysis of the glucose concentration in the sample. Sodium oxalate, EDTA, and heparin are also anticoagulants but do not have the same inhibitory effect on glycolysis as sodium fluoride.Rate this question:
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- 18.
8-hydroxyquinoline is added in the reagents for the analysis fo calcium to:
- A.
Bind to calcium to form a complex that chelates to CPC
- B.
Reduce interferences by magnesium ions
- C.
Enhance color reaction of calcium with CPC
- D.
Prevent percipitation of proteins at acid pH
Correct Answer
B. Reduce interferences by magnesium ionsExplanation
8-hydroxyquinoline is added in the reagents for the analysis of calcium to reduce interferences by magnesium ions. Magnesium ions can interfere with the accurate measurement of calcium levels, as they can form complexes with the reagents used for the analysis. By adding 8-hydroxyquinoline, it selectively binds to magnesium ions, preventing them from interfering with the analysis. This allows for a more accurate determination of the calcium levels in the sample.Rate this question:
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- 19.
Mutarotase is added to the reagent for which of the following monosaccharides?
- A.
Galactose
- B.
Glucose
- C.
Mannose
- D.
Fructose
Correct Answer
B. GlucoseExplanation
Mutarotase is added to the reagent for glucose. Mutarotase is an enzyme that catalyzes the interconversion between the α and β anomers of glucose. Glucose exists in equilibrium between the α and β forms, and mutarotase helps to accelerate this conversion. Therefore, when testing for glucose using a reagent, mutarotase is added to ensure accurate results by allowing the conversion between the two forms of glucose.Rate this question:
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- 20.
Which of the following sets of results would be consistent with heart?
- A.
Inc. CK, LD5, Inc. AST
- B.
Inc. LD1 CK2 and LD
- C.
Inc. alkaline phosphatase and GGT
- D.
Inc. LD1 and CK3
Correct Answer
B. Inc. LD1 CK2 and LDExplanation
The given set of results, which includes an increase in LD1, CK2, and LD, would be consistent with heart. LD1 is an isoenzyme of lactate dehydrogenase that is found predominantly in the heart, while CK2 is a form of creatine kinase that is also found in the heart. Therefore, an increase in LD1 and CK2 levels, along with an increase in LD levels, suggests cardiac damage or injury.Rate this question:
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- 21.
Which of the following interferes with CK measurement?
- A.
N-acetylcysteine
- B.
Mg
- C.
Calcium
- D.
Phosphate
Correct Answer
C. CalciumExplanation
Calcium interferes with CK measurement because it can form complexes with CK, leading to an inaccurate measurement of CK levels. This interference can occur when calcium ions bind to CK, causing a decrease in the enzymatic activity of CK. Therefore, the presence of high levels of calcium can falsely lower the measured CK levels, making it difficult to accurately assess the patient's CK levels.Rate this question:
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- 22.
Analytical procedures that use enzymes as reagents:
- A.
Requires excess substrate
- B.
Add the enzyme(s) in excess
- C.
Must be read every minute for 5 to 10 minute period
- D.
Usually use protein percipitating reagents to stop the reaction
Correct Answer
B. Add the enzyme(s) in excessExplanation
Enzymes are catalysts that speed up chemical reactions. When using enzymes as reagents in analytical procedures, it is necessary to add them in excess to ensure that the reaction proceeds efficiently. This is because enzymes can be used repeatedly and are not consumed in the reaction. By adding enzymes in excess, it ensures that all the substrate molecules are converted into products. This allows for accurate measurement and analysis of the reaction. The other statements provided in the question are not relevant to the explanation of the correct answer.Rate this question:
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- 23.
An absolute ion requirement for alkaline phosphatase analysis is:
- A.
Magnesium, zinc
- B.
Iron, magnesium
- C.
Copper
- D.
Pyridoxyl phosphate
Correct Answer
A. Magnesium, zincExplanation
Magnesium and zinc are absolute ion requirements for alkaline phosphatase analysis. These ions are essential for the proper functioning of alkaline phosphatase enzyme. Magnesium acts as a cofactor for the enzyme, while zinc is involved in maintaining the enzyme's structural integrity. Without these ions, alkaline phosphatase would not be able to perform its enzymatic activity effectively. Therefore, the presence of magnesium and zinc is necessary for accurate analysis of alkaline phosphatase.Rate this question:
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- 24.
Transamination of an ammino acide results in the:
- A.
Formation of ammonia
- B.
Formation of a new amino acid and a anew keto acid
- C.
Formation of urea
- D.
Formation of ammonia, carbon dioxide and water
Correct Answer
B. Formation of a new amino acid and a anew keto acidExplanation
Transamination is a process in which an amino group is transferred from an amino acid to a keto acid, resulting in the formation of a new amino acid and a new keto acid. This process is catalyzed by enzymes called transaminases. Therefore, the correct answer is "formation of a new amino acid and a new keto acid."Rate this question:
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- 25.
Which of the following is the primary reagent used in the Jaffe reaction for creatinine?
- A.
Alkaline copper sulfate
- B.
Phosphotungstic acid
- C.
Diacetyl monoxime
- D.
Alkaline picric acid
Correct Answer
D. Alkaline picric acidExplanation
The Jaffe reaction is a commonly used method for measuring creatinine levels in biological samples. In this reaction, creatinine reacts with alkaline picric acid to form a red-orange colored complex. The intensity of the color is directly proportional to the concentration of creatinine present in the sample. Therefore, alkaline picric acid is the primary reagent used in the Jaffe reaction for creatinine.Rate this question:
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- 26.
The major nonprotein nitrogen degradation product of endogenous purines is
- A.
Urea
- B.
Creatinine
- C.
Picric acid
- D.
Ammonia
Correct Answer
C. Picric acid -
- 27.
In the measurement of urea, urea is initially hydrolyzed by urease to from ammonium ion. The spectrophotometric measurement of the resulting ammonia is reffered to as the
- A.
Ammonia-selective electrode method
- B.
Berthelot reaction
- C.
Jaffe reaction
- D.
Urea method
Correct Answer
B. Berthelot reactionExplanation
The correct answer is Berthelot reaction. In the Berthelot reaction, urea is hydrolyzed by urease to form ammonium ion. The resulting ammonia is then measured using a spectrophotometric method. This reaction is commonly used for the measurement of urea levels in biological samples. The reaction involves the formation of a blue-colored complex when ammonia reacts with hypochlorite and phenol in an alkaline medium. The intensity of the color is directly proportional to the concentration of urea in the sample, allowing for accurate measurement.Rate this question:
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- 28.
Fluoride ion is added to the reagent in the test for CK to inhibit the inhibition from?
- A.
Glutathione
- B.
Magnesium
- C.
Calcium
- D.
Adenylate kinase
Correct Answer
D. Adenylate kinaseExplanation
Fluoride ion is added to the reagent in the test for CK to inhibit the inhibition from adenylate kinase. Adenylate kinase is an enzyme that can interfere with the measurement of creatine kinase (CK) activity. By adding fluoride ion to the reagent, it helps to prevent the inhibition caused by adenylate kinase, allowing for a more accurate measurement of CK activity. The presence of fluoride ion helps to specifically target and inhibit adenylate kinase, ensuring that its interference does not affect the test results.Rate this question:
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- 29.
Which of the following reagents is used in the direct test for the measurement of urea?
- A.
Diacetyl monoxime
- B.
Urease
- C.
Hypochlorite
- D.
Hydroxylamine
Correct Answer
A. Diacetyl monoximeExplanation
Diacetyl monoxime is used in the direct test for the measurement of urea. This reagent reacts with urea to form a yellow-colored complex, which can be measured spectrophotometrically. This method is commonly used in clinical laboratories to determine urea levels in blood or urine samples. Urease, hypochlorite, and hydroxylamine are not used in the direct test for urea measurement.Rate this question:
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- 30.
Creatininase reactions for creatinine assessment
- A.
Yield sarcosine and urea
- B.
Requires the addition of potassium ferricyanide to reduce interferences
- C.
Catalyze the conversion of creatinine to N-methylhydantoin and ammonia
- D.
Catalyze the conversion of creatinine to creatine
Correct Answer
D. Catalyze the conversion of creatinine to creatineExplanation
Creatininase is an enzyme that catalyzes the conversion of creatinine to creatine. Creatinine is a waste product produced by muscles and excreted by the kidneys, while creatine is an important molecule involved in energy metabolism. Therefore, the correct answer states that creatininase reactions catalyze the conversion of creatinine to creatine. This conversion is significant because creatine can be used as a source of energy in various tissues, including the brain and muscles.Rate this question:
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- 31.
Uricase is used in the assay of uric acid to
- A.
Decarboxylate uric acid to form tungsten blue
- B.
Oxidize uric acid to form allantoin
- C.
Reduce uric acid to form allantoin
- D.
Reduce urice acid to form tungsten blue
Correct Answer
B. Oxidize uric acid to form allantoinExplanation
Uricase is an enzyme that catalyzes the oxidation of uric acid to allantoin. Allantoin is a more soluble and less toxic compound compared to uric acid. Therefore, by oxidizing uric acid to allantoin, uricase helps in the detoxification and elimination of uric acid from the body. This is why uricase is used in the assay of uric acid to measure its levels in biological samples.Rate this question:
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- 32.
Blood is obtained in a plain grey top tube containing no serum separator and is transported to the laboratory from an outpatient clinic. Transport time at RT, accessioning and sample prep time equal approx. 3 hrs. If a glucose level is requested on this tube of blood, how might the results be affected by the timing?
- A.
The glucose value would be increased by about 10%
- B.
The glucose value would be decreased bty about 15%
- C.
There would be no effect on the glucose value
- D.
It is no possible to determine from the information given how the glucose value would be affected
Correct Answer
C. There would be no effect on the glucose valueExplanation
The glucose value would not be affected by the timing because glucose is stable in blood for several hours, even without a serum separator. Therefore, the transport time and sample prep time of 3 hours would not impact the glucose value.Rate this question:
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- 33.
In the L to P reaction for measuring LD LD L-lactate + NAD+ <-----> pyruvate + NADH + H +
- A.
The decrease in absorbance of NADH is measured
- B.
The increase in absorbance of NAD+ is measured
- C.
The increase in absorbance of NADH is measured
- D.
NAD+ is measured at 340nm
Correct Answer
C. The increase in absorbance of NADH is measuredExplanation
In the L to P reaction for measuring LD, the decrease in absorbance of NADH is measured. This is because NADH is oxidized to NAD+ during the reaction, resulting in a decrease in its absorbance. By measuring this decrease in absorbance, we can determine the activity of lactate dehydrogenase (LD) in the sample being tested.Rate this question:
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- 34.
The cofactor required in the method of ALT and AST is?
- A.
Pyridoxyl-5-phosphate
- B.
Calcium
- C.
Magnesium
- D.
Phosphate
Correct Answer
A. Pyridoxyl-5-phosphateExplanation
The cofactor required in the method of ALT and AST is pyridoxyl-5-phosphate. This cofactor is also known as vitamin B6 and is essential for the proper functioning of these enzymes. It plays a crucial role in the transfer of amino groups and is necessary for the metabolism of proteins. Without pyridoxyl-5-phosphate, ALT and AST enzymes would not be able to carry out their functions effectively. Calcium, magnesium, and phosphate are not directly involved in the method of ALT and AST.Rate this question:
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- 35.
From the following list, which enzyme transfers a phosphate group?
- A.
GGT
- B.
LD
- C.
CK
- D.
ALT
Correct Answer
C. CKExplanation
CK, also known as creatine kinase, is an enzyme that transfers a phosphate group from creatine phosphate to ADP, forming ATP. This process is essential for the rapid regeneration of ATP, the main energy source for muscle contraction. Therefore, CK is responsible for transferring a phosphate group and plays a crucial role in energy metabolism. GGT (gamma-glutamyltransferase), LD (lactate dehydrogenase), and ALT (alanine transaminase) are enzymes involved in different biochemical reactions and do not transfer phosphate groups.Rate this question:
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- 36.
The enzymatic product commonly associated with oxidase enzymes such as cholesterol oxidase is:
- A.
H2O
- B.
H2O2
- C.
H2O4
- D.
H+
Correct Answer
B. H2O2Explanation
Oxidase enzymes, including cholesterol oxidase, catalyze oxidation reactions. The enzymatic product commonly associated with oxidase enzymes is hydrogen peroxide (H2O2). This is because oxidase enzymes transfer electrons from a substrate to molecular oxygen, resulting in the production of hydrogen peroxide as a byproduct. Therefore, H2O2 is the correct answer.Rate this question:
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- 37.
In the ususal method of HDL cholesterol analysis, the initial step involves
- A.
Hydrolysis of cholesterol esters by cholesterol oxidase to form free cholesterol
- B.
Hydrolysis of cholesterol esters by lipase to form glycerol and free fatty acids
- C.
Precipitation of non-HDl lipoproteins
- D.
Measurement of the rate of oxygne consumption
Correct Answer
C. Precipitation of non-HDl lipoproteinsExplanation
The correct answer is precipitation of non-HDL lipoproteins. In the usual method of HDL cholesterol analysis, the initial step involves precipitating non-HDL lipoproteins from the blood sample. This is done to separate the HDL cholesterol from other lipoproteins in order to accurately measure its levels. Non-HDL lipoproteins include LDL (low-density lipoprotein) and VLDL (very low-density lipoprotein), which are considered to be "bad" cholesterol. By precipitating these lipoproteins, the focus can be solely on measuring the HDL cholesterol levels, which is considered to be "good" cholesterol.Rate this question:
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- 38.
Bilirubin
- A.
In the unconjugated form id highly toxic to the nervous system when elevated
- B.
In the plasma is bound to albumin before conjugation
- C.
Becomes conjugated to hydrochloric acid in hepatocytes
- D.
In the conjugated form is hydrolyzed by beta-glucoronidase from the liver
Correct Answer
B. In the plasma is bound to albumin before conjugationExplanation
Bilirubin in the unconjugated form is highly toxic to the nervous system when elevated. In order to prevent its toxicity, bilirubin is bound to albumin in the plasma before it undergoes conjugation. Conjugation of bilirubin occurs in hepatocytes, where it becomes conjugated to glucuronic acid. The conjugated form of bilirubin is then excreted into the bile and ultimately eliminated from the body. Therefore, the correct answer is that bilirubin in the plasma is bound to albumin before conjugation.Rate this question:
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- 39.
In the initial step of the diazo reaction for total bilirubin determination
- A.
The absorption of light by bilirubin near 460nm is measured
- B.
Bilirubin fractions are oxidized to biliverdin, which id further oxidized to purple
- C.
Serum is added to an aqueous solution of caffeine, sodium acetate and sodium benzoate
- D.
Large quantities of ascorbic acid worsen the detection limit
Correct Answer
C. Serum is added to an aqueous solution of caffeine, sodium acetate and sodium benzoate -
- 40.
What is required of a protein before it can be analyzed by the biuret method?
- A.
Amino and carboxyl termini
- B.
At least two peptide bonds
- C.
Copper moietis
- D.
Quaternary structures
Correct Answer
B. At least two peptide bondsExplanation
Before a protein can be analyzed by the biuret method, it must have at least two peptide bonds. The biuret method is a chemical test that detects the presence of peptide bonds in a substance. Peptide bonds are essential for the formation of proteins, as they link amino acids together to form the protein chain. Therefore, a protein must have at least two peptide bonds in order for the biuret method to accurately detect its presence.Rate this question:
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- 41.
The addition of trichloroacetic acid for turbidimetric protein determination of a CSF specimen is done to
- A.
Dilute the protein in the specimen as it is typically high and beyond the range of standards
- B.
Remove interfering substances from the specimen
- C.
Increase the volume of the specimen because it is difficult to obtain adequate volume
- D.
Precipitate the protein in the specimen
Correct Answer
D. Precipitate the protein in the specimenExplanation
Trichloroacetic acid is added to the CSF specimen for turbidimetric protein determination in order to precipitate the protein in the specimen. This is because turbidimetric protein determination involves measuring the turbidity or cloudiness of the specimen, which is caused by the presence of proteins. By adding trichloroacetic acid, the proteins in the specimen are precipitated or coagulated, making the specimen clearer and reducing the turbidity. This allows for more accurate measurement of protein levels in the specimen.Rate this question:
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- 42.
If you were asked by your lab director to set up an enzyme profile for your chemistry analyzer specifically to assess hepatobiliary disease, which on of the following sets of enzymes would be most appropriate?
- A.
ALT, AST, ALP and GGT
- B.
ALP, GGT and NTP
- C.
Amylase, lipase and GGT
- D.
LD, CK, ALT adn AST
Correct Answer
A. ALT, AST, ALP and GGTExplanation
ALT, AST, ALP, and GGT are all enzymes that can provide valuable information about hepatobiliary disease. ALT (alanine aminotransferase) and AST (aspartate aminotransferase) are liver enzymes that can indicate liver damage or inflammation. ALP (alkaline phosphatase) is an enzyme that can be elevated in conditions affecting the liver and bile ducts. GGT (gamma-glutamyl transferase) is an enzyme that is specifically elevated in liver and bile duct diseases. Therefore, this set of enzymes would be most appropriate for assessing hepatobiliary disease.Rate this question:
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- 43.
A notably hemolyzed serum specimen is received in the lab with the request for CK and LD enzymes and LD isoenzyme electrophoresis. How will the results be falsely affected byt the condition of the specimen?
- A.
LD decreased, CK inc. and inc LD-3
- B.
Inc. LD, CK and LD-1
- C.
Inc LD, CK and LD-2
- D.
Inc. LD and LD-5 with dec. CK
Correct Answer
B. Inc. LD, CK and LD-1Explanation
The hemolysis of the serum specimen can falsely affect the results of the LD and CK enzymes as well as the LD-1 isoenzyme. Hemolysis can release intracellular enzymes, including LD and CK, into the serum, leading to increased levels of these enzymes. LD-1 is the predominant isoenzyme found in red blood cells, so its levels will also be increased in a hemolyzed specimen. Therefore, the correct answer is that the LD, CK, and LD-1 levels will be increased.Rate this question:
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- 44.
A tray of filled uncovered sample cups sits in the laboratory benchtop for an extended period of time. What analytes might be affected if the sample sups are transaprent?
- A.
Protein
- B.
Cholesterol
- C.
Glucose
- D.
Bilirubin
Correct Answer
D. BilirubinExplanation
If the sample cups are transparent and left uncovered for an extended period of time, it is likely that the analyte affected would be bilirubin. Bilirubin is a pigment that is sensitive to light and can degrade when exposed to light for a prolonged period. Therefore, if the sample cups are transparent and not protected from light, the bilirubin levels in the samples may decrease, leading to inaccurate test results.Rate this question:
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- 45.
In carcinoma of the prostate, which of the following enzymes are most likely increased?
- A.
AlkP
- B.
LD5
- C.
Acid phosphatase
- D.
CKMM
Correct Answer
C. Acid phosphataseExplanation
In carcinoma of the prostate, acid phosphatase is most likely increased. Acid phosphatase is an enzyme that is normally found in high levels in the prostate gland. In cases of prostate cancer, the cancer cells can release this enzyme into the bloodstream, leading to elevated levels of acid phosphatase. Therefore, an increase in acid phosphatase levels can be indicative of prostate cancer.Rate this question:
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- 46.
The direct method for the determination of uric acid involves which of the following reagents?
- A.
Ammonium molybdate
- B.
Uricase
- C.
Phosphotungstic acid
- D.
Ammonium phosphate
Correct Answer
C. Phosphotungstic acidExplanation
Phosphotungstic acid is used as a reagent in the direct method for the determination of uric acid. This method involves the reaction between uric acid and phosphotungstic acid, which results in the formation of a blue color complex. The intensity of the color is directly proportional to the concentration of uric acid present in the sample. Therefore, phosphotungstic acid is an essential reagent in this method for the accurate measurement of uric acid levels.Rate this question:
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- 47.
Which of the following reagents id used to precipitate out non-HDL cholesterol?
- A.
Heparin sulfate, manganese chloride
- B.
Heparin sulfate
- C.
Manganese chloride
- D.
Heparin, manganese chloride
Correct Answer
A. Heparin sulfate, manganese chlorideExplanation
Heparin sulfate and manganese chloride are used together as reagents to precipitate out non-HDL cholesterol. Non-HDL cholesterol includes all the cholesterol in the blood except for the HDL (high-density lipoprotein) cholesterol. The combination of heparin sulfate and manganese chloride helps in separating non-HDL cholesterol from other lipoproteins in the blood, allowing for its quantification and analysis.Rate this question:
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- 48.
To reduce interferences in the measurement of albumin using the anionic dyes
- A.
The absorbance is measured 10 mins after the incubation
- B.
The absorbance is measured 30seconds after the initiation of the reaction
- C.
There are no interferences with the anionic dye methods, they are specific for albumin
- D.
Normal end-point reaction
Correct Answer
B. The absorbance is measured 30seconds after the initiation of the reactionExplanation
The correct answer is "the absorbance is measured 30 seconds after the initiation of the reaction" because measuring the absorbance at this specific time allows for accurate measurement of the albumin concentration without interference from other substances. This timing ensures that the reaction has progressed enough to provide a reliable measurement, but is not too far along to introduce additional interferences.Rate this question:
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- 49.
Which of the following bilirubin isomers does not directly react with the Diazo reagent?
- A.
Alpha-bilirubin
- B.
Beta-bilirubin
- C.
Gamma-bilirubin
Correct Answer
A. Alpha-bilirubinExplanation
Alpha-bilirubin does not directly react with the Diazo reagent. This is because the Diazo reagent specifically reacts with the beta-isomer of bilirubin, forming a colored compound. The alpha-isomer does not have the necessary functional groups for this reaction to occur. Therefore, alpha-bilirubin remains unaffected by the Diazo reagent.Rate this question:
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- 50.
Coagulation enzymes synthesized by the liver are an example of?
- A.
Sectretory enzyes with no function in circulation
- B.
Plasma specific enzymes with no function in circulation
- C.
Intracellular enzymes released due to tissue damage
- D.
Plasma specific enzymes with function
Correct Answer
D. Plasma specific enzymes with functionExplanation
The coagulation enzymes synthesized by the liver are plasma specific enzymes with function. These enzymes play a crucial role in the process of blood clotting, which is essential for preventing excessive bleeding. They are specifically produced and released into the plasma to initiate and regulate the coagulation cascade.Rate this question:
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Apr 02, 2024Quiz Edited by
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Dec 08, 2012Quiz Created by
J1_black
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