Quiz : Biochemistry Lab Questions And Answers

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  • 1/77 Questions

    Changed in the presence of noncompetitive inhibitor 

    • Km
    • Vmax
    • Ki
    • None of the above
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We welcome you to these super fun biochemistry lab quiz with well-reserached questions and answers. Biochemistry is a super exciting subject that needs dedication and effort. Why don't you test your basics of the subject with this quiz? All the questions are designed in such a way that will make you think! Do you think you can answer all the See morequestions in our quiz easily? Let's see if you give it a try! Play this quiz with some of your friends to compare scores! Wouldn't that be super cool? We wish you all the very best with this quiz!

Quiz : Biochemistry Lab Questions And Answers - Quiz

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  • 2. 

    The presence of glycerol in sample buffer increasese the density of protein sample and facilitates to prevent from diffision in the sample well 

    • True

    • False

    Correct Answer
    A. True
    Explanation
    The presence of glycerol in the sample buffer increases the density of the protein sample. This increased density helps prevent the proteins from diffusing out of the sample well during electrophoresis. Therefore, the statement is true.

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  • 3. 

    One international unit of an enzyme is defined as the amount that catalyzes: 

    • The formation of one millimol of product in one min

    • The formation of one micromol of product in one min

    • The formation of one mol of product in one min

    • The formation of one micromol of product in one hour

    Correct Answer
    A. The formation of one micromol of product in one min
    Explanation
    An international unit of an enzyme is defined as the amount that catalyzes the formation of one micromol of product in one minute. This means that the enzyme is able to convert one micromol of substrate into product within a time frame of one minute. This unit is commonly used to measure the activity of enzymes and allows for standardized comparisons between different enzyme preparations.

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  • 4. 

    Most of your blood cholesterol is produced by: 

    • Pancreas

    • Heart

    • Liver

    • Kidney

    Correct Answer
    A. Liver
    Explanation
    The liver is responsible for producing most of the blood cholesterol in the body. It synthesizes cholesterol and releases it into the bloodstream to be transported to various tissues. The pancreas, heart, and kidney do not have a significant role in cholesterol production.

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  • 5. 

    The term "blotting" refers to:

    • Running the gel in the presence of SDS

    • Digesting the DNA with restriction enzyme

    • Transfer of biomolecules to an immobilizing membrane

    • Disrupting the disulfide bonds of proteins in the presence of (beta) mercaptoethanol

    Correct Answer
    A. Transfer of biomolecules to an immobilizing membrane
    Explanation
    The term "blotting" refers to the transfer of biomolecules to an immobilizing membrane. This technique is commonly used in molecular biology to separate and identify specific proteins or nucleic acids from a mixture. The biomolecules are first separated by size using gel electrophoresis and then transferred onto a membrane, such as nitrocellulose or PVDF. The transferred biomolecules can then be detected using specific probes, such as antibodies or nucleic acid probes, allowing for further analysis and identification.

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  • 6. 

    Separation of proteins by SDS-electrophoreiss takes adavantage of differences in: 

    • Amino acid composition of proteins

    • Protein isoelectric points

    • Protein size

    • Protein shape

    • Protein solubility

    Correct Answer
    A. Protein size
    Explanation
    SDS-electrophoresis is a technique used to separate proteins based on their size. It involves denaturing the proteins with SDS (sodium dodecyl sulfate) which coats them with a negative charge, causing them to migrate towards the positive electrode. Since the migration rate depends on the size of the protein, smaller proteins move faster and travel further through the gel than larger proteins. Therefore, the separation of proteins by SDS-electrophoresis is based on the differences in protein size.

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  • 7. 

    Most proteins have postive charge above pH 7.0

    • True

    • False

    Correct Answer
    A. False
    Explanation
    Most proteins have a net charge that is dependent on the pH of their environment. At a pH above 7.0, proteins tend to have a negative charge due to the deprotonation of their amino acid side chains. Therefore, the statement that most proteins have a positive charge above pH 7.0 is false.

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  • 8. 

    A Lineweaver-burk (double reciprocal) analysis allows for the determination of: 

    • Km only

    • Vmax only

    • Km and Vmax

    • The reaction product

    • None is true

    Correct Answer
    A. Km and Vmax
    Explanation
    A Lineweaver-Burk analysis is a graphical method used to determine the values of Km (Michaelis constant) and Vmax (maximum velocity) in enzyme kinetics. It involves plotting the reciprocal of the initial reaction rate (1/V0) against the reciprocal of the substrate concentration (1/[S]). The slope of the line represents Km/Vmax, while the y-intercept represents 1/Vmax. By analyzing the Lineweaver-Burk plot, both Km and Vmax can be determined accurately.

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  • 9. 

    Enzymes increase the rate of reaction by:

    • Increasing the free energy of activation

    • Increasing the free energy change of the reaction

    • Decreasing the nergy of activation

    • None of the above

    Correct Answer
    A. Decreasing the nergy of activation
    Explanation
    Enzymes increase the rate of reaction by decreasing the energy of activation. The energy of activation is the energy required for a chemical reaction to occur. By lowering this energy barrier, enzymes allow the reaction to proceed more easily and quickly. This is achieved by providing an alternative pathway with a lower activation energy, allowing more reactant molecules to reach the transition state and form products. Therefore, enzymes effectively decrease the energy of activation and enhance the rate of the reaction.

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  • 10. 

    Which of the following is the proper component sequence in western blotting

    • Fiber pad, filter paper, nitrocellulose membrane, gel filter paper, fiber pad

    • Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad

    • Fiber pad, filter paper, gel, nitrocelluloes membrane, filter paper, fiber pad

    • Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad

    Correct Answer
    A. Fiber pad, filter paper, gel, nitrocelluloes membrane, filter paper, fiber pad
    Explanation
    The proper component sequence in western blotting is as follows: fiber pad, filter paper, gel, nitrocellulose membrane, filter paper, fiber pad. This sequence allows for the transfer of proteins from the gel onto the nitrocellulose membrane, which is an essential step in western blotting. The fiber pads help to maintain even pressure during the transfer process, while the filter papers act as wicks to draw the transfer buffer through the gel and onto the membrane.

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  • 11. 

    The mobility of portein bands in Western blotting is from anode (postive pole) to cathode (negative)

    • True

    • False

    Correct Answer
    A. False
    Explanation
    In Western blotting, the mobility of protein bands is actually from cathode (negative pole) to anode (positive pole). This is because proteins are negatively charged and are attracted towards the positive electrode. Therefore, the statement that the mobility is from anode to cathode is incorrect.

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  • 12. 

    Lysozyme hydrolyses

    • Disulfide bond

    • Phosphodiester bond

    • Hydrogen bond

    • Glycosidic bond

    Correct Answer
    A. Glycosidic bond
    Explanation
    Lysozyme is an enzyme that hydrolyses glycosidic bonds. Glycosidic bonds are a type of bond that connects sugar molecules in carbohydrates. By hydrolyzing these bonds, lysozyme breaks down carbohydrates and helps in the defense against bacterial infections. Therefore, the correct answer is glycosidic bond.

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  • 13. 

    Smaller proteins will have a ______Rf value while larger proteins will have a _____Rf value during gel electrophoresis

    • Smaller, larger

    • Cannot be estimated

    • Larger, smaller

    • None of the above

    Correct Answer
    A. Larger, smaller
    Explanation
    During gel electrophoresis, proteins are separated based on their size and charge. Smaller proteins can move more easily through the gel matrix and therefore have a larger Rf value, indicating that they have traveled a greater distance from the starting point. On the other hand, larger proteins experience more resistance and are not able to move as far, resulting in a smaller Rf value. Therefore, smaller proteins have a larger Rf value while larger proteins have a smaller Rf value during gel electrophoresis.

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  • 14. 

    Factors affecting the velocity of enzyme catalyzed reaction are: 

    • Enzyme concentration only

    • Substrate concentration

    • PH only

    • Temperature only

    • All of the above

    Correct Answer
    A. All of the above
    Explanation
    The velocity of an enzyme-catalyzed reaction can be affected by various factors including enzyme concentration, substrate concentration, pH, and temperature. Enzyme concentration refers to the amount of enzyme present, which can influence the rate of reaction. Substrate concentration, on the other hand, refers to the amount of substrate available for the enzyme to bind to and catalyze. pH can affect the shape and activity of the enzyme, thus impacting the reaction rate. Temperature can also affect the rate of reaction as enzymes have an optimal temperature at which they work most efficiently. Therefore, all of these factors can affect the velocity of an enzyme-catalyzed reaction.

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  • 15. 

    RNA blot refers to:

    • Western blot

    • Northern blot

    • Southern blot

    • Eastern blot

    Correct Answer
    A. Northern blot
    Explanation
    An RNA blot refers to a technique known as northern blotting. It is used to study gene expression by detecting and analyzing RNA molecules. Similar to a western blot that detects proteins, a northern blot specifically detects RNA molecules. This technique involves separating RNA molecules by size using gel electrophoresis, transferring them onto a membrane, and then hybridizing the membrane with a labeled RNA probe to visualize the target RNA of interest.

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  • 16. 

    Cholesterol can easily be metabilized to CO2 and H2O in humans and excreted 

    • False

    • True

    Correct Answer
    A. False
    Explanation
    Cholesterol cannot be easily metabolized to CO2 and H2O in humans and excreted. While some cholesterol can be broken down and eliminated from the body, the majority of it is actually synthesized in the liver and used for various functions in the body. Excess cholesterol can build up in the arteries and lead to health problems such as heart disease. Therefore, it is important to manage cholesterol levels through a healthy diet and lifestyle.

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  • 17. 

    Which of the following expressino is used for determination of Km? 

    • Vo at 1/2 Vmax

    • [S] at 1/2 Vmax

    • [P] at 1/2 Vmax

    • [S] at Vmax

    Correct Answer
    A. [S] at 1/2 Vmax
    Explanation
    The expression [S] at 1/2 Vmax is used for the determination of Km. Km is the Michaelis constant and it represents the substrate concentration at which the reaction velocity is half of the maximum velocity (Vmax). Therefore, measuring the substrate concentration ([S]) at 1/2 Vmax allows us to determine the Km value.

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  • 18. 

    Proteins that are separted in SDS-PAGE are subjected to treat with all of the following reagents/factor EXCEPT: 

    • Tris-buffer

    • Ammonim (APS)

    • B-Mercatoethanol

    • SDS

    • Heat

    Correct Answer
    A. Ammonim (APS)
    Explanation
    SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis) is a technique used to separate proteins based on their molecular weight. In this technique, proteins are denatured and coated with SDS, which imparts a negative charge to all the proteins. The proteins are then subjected to an electric field and migrate towards the positive electrode based on their size. Tris-buffer, b-Mercatoethanol, and heat are all necessary components for SDS-PAGE as they help in protein denaturation, maintaining pH, and breaking disulfide bonds. Ammonium persulfate (APS) is a reagent used to initiate the polymerization of acrylamide in the gel, and it is not directly involved in protein separation.

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  • 19. 

    Name the chemical which is extremely toxic in gel electrophoresis, causing central nervous system paralysis upon absorbing through skin 

    • Ammonium persulfate

    • Acrylamide

    • TEMED

    • SDS

    • None of the above 

    Correct Answer
    A. Acrylamide
    Explanation
    Acrylamide is the correct answer because it is a highly toxic chemical that is commonly used in gel electrophoresis. It can be absorbed through the skin and has the potential to cause central nervous system paralysis. Ammonium persulfate, TEMED, and SDS are not known to be extremely toxic in gel electrophoresis.

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  • 20. 

    During acid phosphatase kinetic analysis, which one of the follwing was used as a substare? 

    • NaCl

    • Bacterial cell wall

    • PNPP

    • Muramic acid

    Correct Answer
    A. PNPP
    Explanation
    pNPP (p-nitrophenyl phosphate) was used as a substrate during acid phosphatase kinetic analysis. This is because acid phosphatase is an enzyme that catalyzes the hydrolysis of phosphate esters under acidic conditions. pNPP is a commonly used substrate for acid phosphatase assays as it can be converted into p-nitrophenol by the enzyme, which can be measured spectrophotometrically. The rate of p-nitrophenol formation is directly proportional to the activity of acid phosphatase, allowing for the determination of enzyme kinetics.

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  • 21. 

    Which of the follwing related to the mchanism of enzymatic reaction 

    • Free energy of activation increases

    • The energy of activation increases

    • Enzymes change the equilibrium constants of the reaction

    • Enzymes decrease the energy of activation

    • Enzymes have no effect in enzyme catalyzed reaction

    Correct Answer
    A. Enzymes decrease the energy of activation
    Explanation
    Enzymes decrease the energy of activation in enzyme-catalyzed reactions. Enzymes are biological catalysts that speed up chemical reactions by lowering the energy barrier, or activation energy, required for the reaction to occur. They achieve this by binding to the reactant molecules, bringing them closer together, and providing an alternative reaction pathway with a lower energy of activation. This allows the reaction to proceed more quickly and efficiently. Therefore, enzymes play a crucial role in facilitating biochemical reactions in living organisms.

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  • 22. 

    Which statement is true for an enzyme catalyzed reaction if [S] >> km 

    • The enzyme activity in the linear range

    • The enzyme is saturated with substrate

    • The enzyme activity is half of the maximal rate

    • The enzyme is showing first order of reaction

    Correct Answer
    A. The enzyme is saturated with substrate
    Explanation
    When [S] (substrate concentration) is much greater than Km (Michaelis constant), it indicates that the enzyme is saturated with substrate. This means that all the enzyme's active sites are occupied by substrate molecules, and any further increase in substrate concentration will not have a significant effect on the reaction rate. Thus, the enzyme activity remains constant at its maximum rate, and the reaction is said to be in the saturated state.

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  • 23. 

    The use of methanol in transfer buffer of western blotting is

    • To increase negative charge of proteins to be transferred

    • To dissociate the SDS from proteins and facilitate to move from gel to NC membrane

    • To increase the dielectric constant

    • Methanol has nothing to do with western blotting

    Correct Answer
    A. To dissociate the SDS from proteins and facilitate to move from gel to NC membrane
    Explanation
    Methanol is used in the transfer buffer of western blotting to dissociate the SDS (sodium dodecyl sulfate) from proteins. SDS is commonly used in protein denaturation and electrophoresis, but it can interfere with the transfer of proteins from the gel to the nitrocellulose (NC) membrane. Methanol helps to remove the SDS, allowing the proteins to move more easily from the gel to the NC membrane during the transfer process.

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  • 24. 

    Which one of the following does not apply to Michaelis-Menten assumptions: 

    • The E, S, and ES are in rapid equilibrium

    • The product formation is inversely proportional to ES complex concentration

    • The product formation is directly proportional to ES complex concentration

    Correct Answer
    A. The product formation is inversely proportional to ES complex concentration
    Explanation
    The Michaelis-Menten assumptions state that the E, S, and ES (enzyme, substrate, and enzyme-substrate complex) are in rapid equilibrium, and the product formation is directly proportional to the ES complex concentration. However, the assumption that does not apply is that the product formation is inversely proportional to the ES complex concentration.

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  • 25. 

    Blocking step in western blot detection is used to increase nonspecific binding

    • True

    • False

    Correct Answer
    A. False
    Explanation
    Blocking step in western blot detection is used to decrease nonspecific binding. During the blocking step, a solution containing proteins such as bovine serum albumin (BSA) or nonfat dry milk is applied to the membrane. This solution fills up any unbound sites on the membrane, preventing nonspecific binding of antibodies or other detection reagents. By blocking nonspecific binding, the blocking step allows for more specific and accurate detection of the target protein of interest. Therefore, the correct answer is False.

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  • 26. 

    Wht is the v/Vmax ratio if [S]=3Km?

    • 75

    • 0.75

    • 0.25

    • None

    Correct Answer
    A. 0.75
    Explanation
    The v/Vmax ratio represents the fraction of the maximum velocity (Vmax) that is achieved at a given substrate concentration ([S]). In this case, the substrate concentration is given as 3 times the Michaelis constant (Km). The Michaelis constant represents the substrate concentration at which the reaction rate is half of the maximum velocity. Therefore, if [S]=3Km, it means that the substrate concentration is three times higher than the concentration at which the reaction rate is half of Vmax. As a result, the v/Vmax ratio would be 0.75, indicating that the reaction is operating at 75% of its maximum velocity.

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  • 27. 

    Which property of proten best determines the electrophoretic pattern in SDS-PAGe under reducing conditions? 

    • Shape of the native protein

    • Molecular weight of subunits

    • Specific binding sites on the native proten

    • Net charge of the native protein

    • All of the above

    Correct Answer
    A. Molecular weight of subunits
    Explanation
    The electrophoretic pattern in SDS-PAGE under reducing conditions is primarily determined by the molecular weight of subunits. SDS-PAGE separates proteins based on their molecular weight, and reducing conditions break disulfide bonds, resulting in proteins being denatured and separated into subunits. The smaller subunits will migrate faster through the gel, while larger subunits will migrate slower, resulting in distinct bands on the gel. The other properties mentioned, such as the shape of the native protein, specific binding sites, and net charge, may also contribute to the electrophoretic pattern to some extent, but the molecular weight of subunits is the main determining factor.

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  • 28. 

    One international unit of an enzyme is defined as the amount that catalyzes:

    • The formation of one mmol of product in one min

    • The formation of one micromol of product in one min

    • The formation of one mol of product in one min

    • The formation of one micromol of product in one hour

    Correct Answer
    A. The formation of one micromol of product in one min
    Explanation
    An international unit of an enzyme is a standardized measure of its activity. In this case, it is defined as the amount of enzyme that catalyzes the formation of one micromol of product in one minute. This means that if the enzyme is able to produce one micromol of product in one minute, it is considered to have one international unit of activity. This definition allows for a standardized way of comparing the activity of different enzymes.

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  • 29. 

    The polymerized gel is an ideal gel matrix that provides desired prosity and is toxic after polymerization 

    • True

    • False

    Correct Answer
    A. False
    Explanation
    The statement is false because a polymerized gel is not toxic after polymerization. Polymerization is the process of converting a liquid monomer into a solid polymer network. Once the gel is polymerized, it becomes a stable and non-toxic material. The polymerized gel can be used as an ideal gel matrix for various applications due to its desired porosity and stability.

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  • 30. 

    Following are the properties of Lysozyme except:

    • Has MW of 14,000 dalton

    • Works on Micrococcus lysodeikticus cell wall

    • Hydrolyses phosphodiester bond

    • Provides defense mechanism against bacterial infection

    • Rich in tears

    Correct Answer
    A. Hydrolyses phosphodiester bond
    Explanation
    Lysozyme is an enzyme that works on the cell wall of Micrococcus lysodeikticus, hydrolyzing the glycosidic bond between N-acetylglucosamine and N-acetylmuramic acid. This enzymatic activity is responsible for its ability to provide a defense mechanism against bacterial infection. Lysozyme is also found to be rich in tears. However, it does not hydrolyze phosphodiester bonds.

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  • 31. 

    CHolesterol is synthesized from 

    • Choline

    • Lipoic acid

    • Acetyl CoA

    • Malic acid

    Correct Answer
    A. Acetyl CoA
    Explanation
    Acetyl CoA is the precursor for cholesterol synthesis. It is formed from the breakdown of glucose or fatty acids in the mitochondria. Acetyl CoA is then converted into mevalonate, which is the first step in the synthesis of cholesterol. Choline, lipoic acid, and malic acid are not directly involved in the synthesis of cholesterol.

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  • 32. 

    Changed in the presence of competitive inhibitor 

    • Km

    • Vmax

    • Ki

    • None of the above

    Correct Answer
    A. Km
    Explanation
    The correct answer is Km. Km is a measure of the affinity between an enzyme and its substrate. In the presence of a competitive inhibitor, the inhibitor competes with the substrate for binding to the active site of the enzyme. This increases the apparent Km value, as more substrate is required to achieve the same reaction rate. The Vmax value, on the other hand, remains unchanged in the presence of a competitive inhibitor. Ki refers to the inhibition constant, which is not directly related to changes in Km. Therefore, the correct answer is Km.

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  • 33. 

    Gives approximate substrate concentration within the cell 

    • Km

    • Vmax

    • Ki

    • None of the above

    Correct Answer
    A. Km
    Explanation
    Km is a measure of the affinity between an enzyme and its substrate. It represents the substrate concentration at which the enzyme achieves half of its maximum velocity (Vmax). Therefore, Km gives an approximate indication of the substrate concentration within the cell. A lower Km value indicates a higher affinity between the enzyme and substrate, meaning that the enzyme can achieve half of its maximum velocity at lower substrate concentrations. Conversely, a higher Km value indicates a lower affinity between the enzyme and substrate, requiring higher substrate concentrations to achieve half of the maximum velocity.

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  • 34. 

    A 0.2 ml pure Muramidase (2.5 mg/ml) hydrolyzed 0.2 mmol of cell wall in 5 minutes. What is the specific activity

    • 40 micromol/min/mg

    • 0.04 millimol/min/mg

    • 80 micromol/min/mg

    • None

    Correct Answer
    A. 80 micromol/min/mg
    Explanation
    The specific activity of an enzyme is a measure of its catalytic efficiency and is typically expressed as the amount of substrate converted per unit time per unit of enzyme concentration. In this question, the enzyme Muramidase hydrolyzed 0.2 mmol of cell wall in 5 minutes, and the concentration of the enzyme is 2.5 mg/ml. To calculate the specific activity, we divide the amount of substrate converted (0.2 mmol) by the time (5 minutes) and the enzyme concentration (2.5 mg/ml). This gives us a specific activity of 80 micromol/min/mg.

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  • 35. 

    Which tracking dye did you use in SDS=PAGE?

    • Coomassie blue

    • Bromophenol blue

    • Bromothymol blue

    • DTT

    Correct Answer
    A. Bromophenol blue
    Explanation
    Bromophenol blue is commonly used as a tracking dye in SDS-PAGE. It is added to the sample before loading onto the gel to monitor the progress of the electrophoresis. The dye migrates at a predictable rate, allowing the researcher to track the movement of the proteins and determine when to stop the electrophoresis. It is easily visible due to its blue color and does not interfere with the separation of the proteins.

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  • 36. 

    A protein mixture having a MW of 2,000 23000 and 40,000 daltons was electrophoresed in SDS-PAGE system. Which protein will be at the bottom of gel? 

    • A protein with MW of 40,000 daltons

    • A protein with MW of 23000 daltons

    • A protein with MW of 2,000 daltons

    • All of them will be in one band

    Correct Answer
    A. A protein with MW of 2,000 daltons
    Explanation
    The protein with a molecular weight (MW) of 2,000 daltons will be at the bottom of the gel in the SDS-PAGE system. In this system, smaller proteins migrate faster and move further down the gel compared to larger proteins. Since the protein with a MW of 2,000 daltons is the smallest among the given options, it will migrate the furthest and be at the bottom of the gel.

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  • 37. 

    Polyacrylamide gels are prepared by polymerization of acrylamide monomer and N-N'-methylene-bis-acrylamide cross-linker in the presence of: 

    • TEMED + b- mercaptoethaol

    • APS + b- mercaptoethaol

    • APS + TEMED

    • Tris- buffer + phosphate ion

    Correct Answer
    A. APS + TEMED
    Explanation
    Polyacrylamide gels are prepared by polymerization of acrylamide monomer and N-N'-methylene-bis-acrylamide cross-linker in the presence of APS (Ammonium persulfate) and TEMED (Tetramethylethylenediamine). APS acts as a initiator for the polymerization reaction by generating free radicals, while TEMED acts as a catalyst to enhance the reaction rate. Both APS and TEMED are commonly used in gel electrophoresis to initiate the formation of the polyacrylamide gel matrix.

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  • 38. 

    Which of the follwing is the caharacteristic (s) of pH curve in enzyme kinetic analysis? 

    • Rectangular

    • Sigmoidal

    • Hyperbolic

    • Linear

    • Bell shape

    Correct Answer
    A. Bell shape
    Explanation
    In enzyme kinetic analysis, the pH curve is bell-shaped. This means that as the pH of the environment changes, the activity of the enzyme also changes. At a certain optimal pH, the enzyme activity is at its maximum. As the pH deviates from this optimal value, the enzyme activity decreases. This bell-shaped curve is a characteristic feature of pH-dependent enzyme kinetics and is often used to study the effect of pH on enzyme activity.

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  • 39. 

    Which of the following serves as a substrate for Acid phosphatase? 

    • PNPP

    • PNP

    • KOH

    • Cell wall

    Correct Answer
    A. PNPP
    Explanation
    pNPP (p-nitrophenyl phosphate) serves as a substrate for Acid phosphatase. Acid phosphatase is an enzyme that catalyzes the hydrolysis of phosphate esters, and pNPP is commonly used as a substrate to measure its activity. When pNPP is hydrolyzed by Acid phosphatase, it produces pNP (p-nitrophenol) as a product. The amount of pNP produced can be measured spectrophotometrically, allowing for the quantification of Acid phosphatase activity. Therefore, pNPP is the correct answer as it specifically serves as a substrate for Acid phosphatase.

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  • 40. 

    In competitive inhibiton, the inhibitor binds to the: 

    • ES complex

    • Free enzyme

    • Both free enzyme and ES complex

    • None of the above

    Correct Answer
    A. Free enzyme
    Explanation
    In competitive inhibition, the inhibitor binds to the free enzyme. This means that the inhibitor competes with the substrate for binding to the active site of the enzyme. When the inhibitor is bound to the enzyme, it prevents the substrate from binding and thus inhibits the enzyme's activity. This type of inhibition can be overcome by increasing the concentration of the substrate, as it increases the chances of the substrate outcompeting the inhibitor for binding to the enzyme.

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  • 41. 

    A prosthetic group 

    • Is composed of polypeptides

    • Does not participate in chemical reactions

    • Is a tightly bound nonprotein part of an enzyme

    • Is present in all enzymes

    • Is a nonprotein enzyme

    Correct Answer
    A. Is a tightly bound nonprotein part of an enzyme
    Explanation
    A prosthetic group is a tightly bound nonprotein part of an enzyme. Prosthetic groups are small molecules that are permanently attached to enzymes and are necessary for their proper functioning. They can be organic molecules such as vitamins or inorganic molecules such as metal ions. Prosthetic groups play a crucial role in enzyme catalysis and can participate in the enzyme's chemical reactions. They are specific to certain enzymes and are not present in all enzymes.

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  • 42. 

    A catalytically inactive protein formed by removal of the cofactor from an active enzyme is referred to as: 

    • Apoenzyme

    • Prosthetic group

    • Coenzyme

    • Holoenzyme

    Correct Answer
    A. Apoenzyme
    Explanation
    When the cofactor is removed from an active enzyme, the resulting protein is called an apoenzyme. The cofactor is a non-protein molecule that is necessary for the enzyme to function properly. Without the cofactor, the enzyme loses its catalytic activity and becomes inactive. The apoenzyme can regain its activity when the cofactor is reattached, forming a holoenzyme. Therefore, the correct answer is apoenzyme.

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  • 43. 

    All of the following chemicals are the components of transfer buffer except

    • Methanol

    • Tris-buffer

    • SDS

    • Glycine

    Correct Answer
    A. Methanol
    Explanation
    All of the following chemicals are the components of transfer buffer except methanol. The other components—tris-buffer, SDS, and glycine—are commonly used in transfer buffers for protein gel electrophoresis. 

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  • 44. 

    Which of the following is the proper component sequence in western blotting: 

    • Fiber pad, filter paper, nitrocellulose membrane, gel, filter paper, fiber pad

    • Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad

    • Fiber pad, filter paper, gel nitrocellulose membrane filter paper, fiber pad

    • Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad

    Correct Answer
    A. Fiber pad, filter paper, gel nitrocellulose membrane filter paper, fiber pad
    Explanation
    The proper component sequence in western blotting is as follows: fiber pad, filter paper, gel, nitrocellulose membrane, filter paper, fiber pad. This sequence ensures that the gel is sandwiched between the filter papers, which helps in the transfer of proteins from the gel to the nitrocellulose membrane. The fiber pads act as wicks to facilitate the transfer of buffer and prevent drying out of the gel.

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  • 45. 

    Which of the following statement(s) is ture about lysozyme enzyme?

    • Provide defense mechanism against bacterial infection

    • Cleaves the glycosidic bond

    • Eggwhite is the source for this enzyme

    • All of the statements are true

    Correct Answer
    A. All of the statements are true
    Explanation
    All of the statements are true. Lysozyme is an enzyme that provides a defense mechanism against bacterial infection by breaking down the glycosidic bond in bacterial cell walls. It is found in various sources, including egg white.

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  • 46. 

    Which of the following method did you use for calculating Ki in the enzyme inhibition lab?

    • Michaelis- Menten equation

    • Dixon plot

    • Eadie-Hofstee plot

    • None of the above

    Correct Answer
    A. Dixon plot
    Explanation
    The Dixon plot is a graphical method used to determine the inhibition constant (Ki) in enzyme inhibition experiments. It involves plotting the reciprocal of the reaction rate against the inhibitor concentration, and the slope of the resulting line can be used to calculate the Ki value. This method is particularly useful for determining the type of inhibition (competitive, non-competitive, or uncompetitive) and provides a visual representation of the data, making it easier to interpret and analyze.

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  • 47. 

    In the last experiment, you added lysozyme just before recording the absorbance reading. At the end of reaction, you noticed clear assay solution. What could be the reason? Is it..

    • The enzyme cleaves the hydrogen bond and disrupts the cell wall that becomes clear

    • The enzyme cleaves the disulfide bond and disrupts the cell wall that becomes clear

    • The enzyme cleaves the glycosidic bond and disrupts the cell wall that becomes clear

    • None of the above

    Correct Answer
    A. The enzyme cleaves the glycosidic bond and disrupts the cell wall that becomes clear
    Explanation
    The clear assay solution observed at the end of the reaction suggests that the enzyme, lysozyme, cleaves the glycosidic bond and disrupts the cell wall. This is because lysozyme is known to have the ability to break down the glycosidic bond found in the peptidoglycan layer of bacterial cell walls. When this bond is cleaved, it weakens the cell wall structure, causing it to become clear. The other options, such as cleaving hydrogen or disulfide bonds, do not have the same effect on the cell wall and would not result in a clear assay solution.

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  • 48. 

    If Vmax = 150 micromol/sec and Km = 2mM, what is the initial velocity at  [S]= 10 mM? 

    • 125 micromoles/sec

    • 91 micromoles/sec

    • 125 micromol/min

    • 300 micromol/sec

    Correct Answer
    A. 125 micromoles/sec
    Explanation
    The initial velocity of a reaction can be calculated using the Michaelis-Menten equation, which is V0 = (Vmax * [S]) / (Km + [S]). In this case, Vmax is given as 150 micromol/sec and Km is given as 2mM. The concentration of substrate, [S], is given as 10 mM. Plugging these values into the equation, we get V0 = (150 * 10) / (2 + 10) = 1500 / 12 = 125 micromoles/sec. Therefore, the initial velocity at [S] = 10 mM is 125 micromoles/sec.

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  • 49. 

    Which of the following binds plasma cholesterol and transport cholesterol to the liver? 

    • LDL

    • Chylomicrons

    • HDL

    • VLDL

    Correct Answer
    A. HDL
    Explanation
    HDL, or high-density lipoprotein, binds plasma cholesterol and transports it to the liver. HDL is often referred to as "good cholesterol" because it helps remove excess cholesterol from the bloodstream and carries it to the liver for processing and excretion. This process helps to prevent the buildup of cholesterol in the arteries and reduces the risk of heart disease.

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Quiz Review Timeline (Updated): Feb 6, 2024 +

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  • Current Version
  • Feb 06, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Jul 22, 2011
    Quiz Created by
    Ekanye

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