Quiz : Biochemistry Lab Questions And Answers
(618).jpg "Quiz : Biochemistry Lab Questions And Answers - Quiz")
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 questions 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!
- 1.
The term "blotting" refers to:
- A.
Running the gel in the presence of SDS
- B.
Digesting the DNA with restriction enzyme
- C.
Transfer of biomolecules to an immobilizing membrane
- D.
Disrupting the disulfide bonds of proteins in the presence of (beta) mercaptoethanol
Correct Answer
C. Transfer of biomolecules to an immobilizing membraneExplanation
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.Rate this question:
-
- 2.
Check all choices that do not belong to transfer buffer:
- A.
Methanol
- B.
Tris-buffer
- C.
SDS
- D.
Acetone
- E.
Glycine
- F.
2-mercaptoethanol
Correct Answer(s)
C. SDS
D. Acetone
F. 2-mercaptoethanolExplanation
The question asks us to identify the choices that do not belong to the transfer buffer. A transfer buffer is a solution used in protein transfer during Western blotting. Methanol, tris-buffer, glycine, and 2-mercaptoethanol are commonly used components of transfer buffers. However, SDS, acetone, and 2-mercaptoethanol are not typically included in transfer buffers. Therefore, SDS, acetone, and 2-mercaptoethanol do not belong to the transfer buffer.Rate this question:
-
- 3.
Which reagent was used for total protein detection in your Western blot lab?
- A.
Ninhydrin spray
- B.
Silver stain
- C.
Colloidal gold reagent
- D.
Copper stain
Correct Answer
C. Colloidal gold reagentExplanation
The colloidal gold reagent was used for total protein detection in the Western blot lab.Rate this question:
-
- 4.
RNA blot refers to:
- A.
Western blot
- B.
Northern blot
- C.
Southern blot
- D.
Eastern blot
Correct Answer
B. Northern blotExplanation
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.Rate this question:
-
- 5.
What are the basic steps required for biological molecules' detection using western blot technique?
- A.
Electrophoretic separation of protein
- B.
Transfer and immobilization onto a membrane
- C.
Binding of probe molecule to the target molecules on the membrane
- D.
Visualization of bound protein
Correct Answer(s)
A. ElectropHoretic separation of protein
B. Transfer and immobilization onto a membrane
C. Binding of probe molecule to the target molecules on the membrane
D. Visualization of bound protein -
- 6.
The mobility of portein bands in Western blotting is from anode (postive pole) to cathode (negative)
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
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.Rate this question:
-
- 7.
Most proteins have postive charge above pH 7.0
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
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.Rate this question:
-
- 8.
Blocking step in western blot detection is used to increase nonspecific binding
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
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.Rate this question:
-
- 9.
Which of the following is the proper component sequence in western blotting:
- A.
Fiber pad, filter paper, nitrocellulose membrane, gel, filter paper, fiber pad
- B.
Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad
- C.
Fiber pad, filter paper, gel nitrocellulose membrane filter paper, fiber pad
- D.
Fiber pad, nitrocellulose membrane, filter paper, gel, filter paper, fiber pad
Correct Answer
C. Fiber pad, filter paper, gel nitrocellulose membrane filter paper, fiber padExplanation
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.Rate this question:
-
- 10.
The use of methanol in transfer buffer of western blotting is
- A.
To increase negative charge of proteins to be transferred
- B.
To dissociate the SDS from proteins and facilitate to move from gel to NC membrane
- C.
To increase the dielectric constant
- D.
Methanol has nothing to do with western blotting
Correct Answer
B. To dissociate the SDS from proteins and facilitate to move from gel to NC membraneExplanation
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.Rate this question:
-
- 11.
Bell shape curve in an enzyme characterization is a represenative of:
- A.
PH only
- B.
Temperature only
- C.
Enzyme concentration only
- D.
A and c
- E.
A and b
Correct Answer
E. A and bExplanation
The bell shape curve in an enzyme characterization is representative of pH and temperature. This is because both pH and temperature can affect the activity and stability of enzymes. Enzymes have an optimal pH and temperature at which they function most efficiently, and deviations from these optimal conditions can lead to a decrease in enzyme activity. Therefore, the bell shape curve demonstrates the relationship between pH and temperature and their impact on enzyme function.Rate this question:
-
- 12.
What will be the inital velocity (Vo) for an enzyme that has Km =2s
- A.
1/3 Vmax
- B.
1/2 Vmax
- C.
2/3 Vmax
- D.
None of the above
Correct Answer
A. 1/3 VmaxExplanation
The correct answer is 1/3 Vmax. This is because the initial velocity (Vo) for an enzyme is equal to 1/3 of the maximum velocity (Vmax). The Km value of 2s does not directly determine the initial velocity, but rather represents the substrate concentration at which the enzyme achieves half of its maximum velocity. Therefore, the initial velocity can be calculated as a fraction of Vmax, with 1/3 being the correct ratio.Rate this question:
-
- 13.
Which of the following statement(s) is ture about lysozyme enzyme?
- A.
Provide defense mechanism against bacterial infection
- B.
Cleaves the glycosidic bond
- C.
Eggwhite is the source for this enzyme
- D.
All of the statements are true
Correct Answer
D. All of the statements are trueExplanation
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.Rate this question:
-
- 14.
One international unit of an enzyme is defined as the amount that catalyzes:
- A.
The formation of one mmol of product in one min
- B.
The formation of one micromol of product in one min
- C.
The formation of one mol of product in one min
- D.
The formation of one micromol of product in one hour
Correct Answer
B. The formation of one micromol of product in one minExplanation
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.Rate this question:
-
- 15.
When [S] <
- A.
The reaction is in zero order of reaction
- B.
The reaction is in first order of reaction
- C.
The product formation is directly related to the substrate concentration
- D.
It is the substrate concentration that equals the Km estimate
- E.
B and c
Correct Answer
E. B and cExplanation
When the substrate concentration is less than the Km (Michaelis constant), the reaction is in the zero order of reaction. In this case, the product formation is not directly related to the substrate concentration. However, when the substrate concentration is equal to or greater than the Km, the reaction is in the first order of reaction and the product formation becomes directly related to the substrate concentration. Therefore, the correct answer is option b and c.Rate this question:
-
- 16.
If Vmax = 150 micromol/sec and Km = 2mM, what is the initial velocity at [S]= 10 mM?
- A.
125 micromoles/sec
- B.
91 micromoles/sec
- C.
125 micromol/min
- D.
300 micromol/sec
Correct Answer
A. 125 micromoles/secExplanation
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.Rate this question:
-
- 17.
Wht is the v/Vmax ratio if [S]=3Km?
- A.
75
- B.
0.75
- C.
0.25
- D.
None
Correct Answer
B. 0.75Explanation
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.Rate this question:
-
- 18.
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..
- A.
The enzyme cleaves the hydrogen bond and disrupts the cell wall that becomes clear
- B.
The enzyme cleaves the disulfide bond and disrupts the cell wall that becomes clear
- C.
The enzyme cleaves the glycosidic bond and disrupts the cell wall that becomes clear
- D.
None of the above
Correct Answer
C. The enzyme cleaves the glycosidic bond and disrupts the cell wall that becomes clearExplanation
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.Rate this question:
-
- 19.
Enzymes increase the rate of reaction by:
- A.
Increasing the free energy of activation
- B.
Increasing the free energy change of the reaction
- C.
Decreasing the nergy of activation
- D.
None of the above
Correct Answer
C. Decreasing the nergy of activationExplanation
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.Rate this question:
-
- 20.
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?
- A.
40 micromole/min/mg
- B.
0.04 millimol/min/mg
- C.
80 micromol/min/mg
- D.
None of the above
Correct Answer
C. 80 micromol/min/mgExplanation
The specific activity of an enzyme is a measure of its catalytic efficiency and is calculated by dividing the rate of reaction (in this case, the hydrolysis of cell wall) by the amount of enzyme present (in this case, the concentration of Muramidase).
In this question, 0.2 ml of Muramidase with a concentration of 2.5 mg/ml hydrolyzed 0.2 mmol of cell wall in 5 minutes.
To calculate the specific activity, we first need to convert the units to be consistent.
The amount of Muramidase used in the reaction is 0.2 ml * 2.5 mg/ml = 0.5 mg.
The rate of reaction is 0.2 mmol / 5 minutes = 0.04 mmol/min.
Therefore, the specific activity is 0.04 mmol/min / 0.5 mg = 80 micromol/min/mg.
Hence, the correct answer is 80 micromol/min/mg.Rate this question:
-
- 21.
Modified by a suitable inhibitor
- A.
Km
- B.
Vmax
- C.
Ki
- D.
None of the above
Correct Answer
C. KiExplanation
The given options are related to enzyme kinetics. Km represents the Michaelis constant, which is a measure of the affinity between the enzyme and substrate. Vmax represents the maximum rate of reaction. Ki represents the inhibition constant, which is a measure of the affinity between the enzyme and an inhibitor. Among the given options, "Modified by a suitable inhibitor" is the most appropriate explanation for Ki, as it indicates that the enzyme's activity is altered by the presence of an inhibitor.Rate this question:
-
- 22.
Changed in the presence of competitive inhibitor
- A.
Km
- B.
Vmax
- C.
Ki
- D.
None of the above
Correct Answer
A. KmExplanation
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.Rate this question:
-
- 23.
Changed in the presence of noncompetitive inhibitor
- A.
Km
- B.
Vmax
- C.
Ki
- D.
None of the above
Correct Answer
B. VmaxExplanation
In the presence of a noncompetitive inhibitor, the Vmax (maximum velocity) of the enzyme-catalyzed reaction is affected. Noncompetitive inhibitors bind to the enzyme or the enzyme-substrate complex at a site other than the active site, causing a change in the enzyme's conformation and reducing its catalytic activity. This results in a decrease in the rate of the reaction, leading to a lower Vmax. The Km (Michaelis constant) is not affected by noncompetitive inhibitors as they do not compete with the substrate for binding to the active site. Therefore, the correct answer is Vmax.Rate this question:
-
- 24.
Changed in the presence of uncompetitive inhibitor
- A.
Km
- B.
Vmax
- C.
Ki
- D.
None of the above
Correct Answer
A. KmExplanation
Km is the Michaelis constant, which represents the substrate concentration at which the reaction rate is half of its maximum velocity (Vmax). In the presence of an uncompetitive inhibitor, the inhibitor binds to the enzyme-substrate complex, preventing the release of the product and reducing the effective concentration of the enzyme-substrate complex. This results in an increase in Km, as higher substrate concentrations are required to reach half of the maximum reaction rate. Therefore, the correct answer is Km.Rate this question:
-
- 25.
Gives approximate substrate concentration within the cell
- A.
Km
- B.
Vmax
- C.
Ki
- D.
None of the above
Correct Answer
A. KmExplanation
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.Rate this question:
-
- 26.
Which of the following is a noncompetitive inhibitor for acid phosphatase?
- A.
Sodium arsenate
- B.
Sodium fluoride
- C.
Sodium phosphate
- D.
PNPP
Correct Answer
B. Sodium fluorideExplanation
Sodium fluoride is a noncompetitive inhibitor for acid phosphatase because it binds to a site on the enzyme that is distinct from the active site. This binding alters the enzyme's conformation and reduces its activity, regardless of the substrate concentration. In contrast, competitive inhibitors bind to the active site and compete with the substrate for binding, while uncompetitive inhibitors only bind to the enzyme-substrate complex. Sodium arsenate and sodium phosphate are not noncompetitive inhibitors for acid phosphatase. pNPP is a substrate for acid phosphatase, not an inhibitor.Rate this question:
-
- 27.
Which of the following method did you use for calculating Ki in the enzyme inhibition lab?
- A.
Michaelis- Menten equation
- B.
Dixon plot
- C.
Eadie-Hofstee plot
- D.
None of the above
Correct Answer
B. Dixon plotExplanation
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.Rate this question:
-
- 28.
In competitive inhibiton, the inhibitor binds to the:
- A.
ES complex
- B.
Free enzyme
- C.
Both free enzyme and ES complex
- D.
None of the above
Correct Answer
B. Free enzymeExplanation
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.Rate this question:
-
- 29.
Which of the following expressino is used for determination of Km?
- A.
Vo at 1/2 Vmax
- B.
[S] at 1/2 Vmax
- C.
[P] at 1/2 Vmax
- D.
[S] at Vmax
Correct Answer
B. [S] at 1/2 VmaxExplanation
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.Rate this question:
-
- 30.
Substances generally form covalent bonds with a specific functional group in an enzyme's active site are categorized as
- A.
Competitive inhibitor
- B.
Uncompetitive inhibitor
- C.
Irreversible inhibitors
- D.
Noncompetitive inhibitors
Correct Answer
C. Irreversible inhibitorsExplanation
Substances that form covalent bonds with a specific functional group in an enzyme's active site are categorized as irreversible inhibitors. Unlike reversible inhibitors, irreversible inhibitors permanently bind to the enzyme, rendering it inactive. This occurs through the formation of a covalent bond between the inhibitor and the enzyme's active site, preventing the enzyme from carrying out its normal function.Rate this question:
-
- 31.
Which property of proten best determines the electrophoretic pattern in SDS-PAGe under reducing conditions?
- A.
Shape of the native protein
- B.
Molecular weight of subunits
- C.
Specific binding sites on the native proten
- D.
Net charge of the native protein
- E.
All of the above
Correct Answer
B. Molecular weight of subunitsExplanation
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.Rate this question:
-
- 32.
Which reagent/ chemical initiate polymerization in gel electrophoresis?
- A.
TEMED
- B.
Ammonium persulfate
- C.
Acrylamide
- D.
2-mercaptoethanol
- E.
SDS
Correct Answer
B. Ammonium persulfateExplanation
Ammonium persulfate is the reagent/chemical that initiates polymerization in gel electrophoresis. It works as a radical initiator, which means it generates free radicals that start the polymerization process. These free radicals react with the monomers, such as acrylamide, to form a polymer matrix, creating the gel. TEMED is often used in conjunction with ammonium persulfate to enhance the polymerization process. Acrylamide is the monomer that forms the gel matrix, while 2-mercaptoethanol is a reducing agent that helps break disulfide bonds in proteins. SDS is a detergent used to denature proteins and make them negatively charged for electrophoretic separation.Rate this question:
-
- 33.
Proteins that are to be separted in SDS-PAGE are subjected to treat with all of the follwing reagents except:
- A.
Heat
- B.
SDS
- C.
B-mercaptoethanol
- D.
Ammonium persulfate
Correct Answer
D. Ammonium persulfateExplanation
Ammonium persulfate is not used to treat proteins in SDS-PAGE. SDS (sodium dodecyl sulfate) is a detergent that denatures proteins and gives them a negative charge. B-mercaptoethanol is a reducing agent that breaks disulfide bonds in proteins. Heat is used to denature proteins. However, ammonium persulfate is not used in the treatment process of proteins in SDS-PAGE.Rate this question:
-
- 34.
Which tracking dye did you use in SDS=PAGE?
- A.
Coomassie blue
- B.
Bromophenol blue
- C.
Bromothymol blue
- D.
DTT
Correct Answer
B. BromopHenol blueExplanation
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.Rate this question:
-
- 35.
The polymerized gel is an ideal gel matrix that provides desired prosity and is toxic after polymerization
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
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.Rate this question:
-
- 36.
The presence of glycerol in sample buffer increasese the density of protein sample and facilitates to prevent from diffision in the sample well
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
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.Rate this question:
-
- 37.
At low pH, TEMED may become deprotonated and results in slower polymerization
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
At low pH, TEMED does not become deprotonated and does not result in slower polymerization. TEMED is a commonly used catalyst in protein gel electrophoresis and is typically added to the gel solution to initiate the polymerization of acrylamide and bis-acrylamide. It acts as a source of free radicals that react with the acrylamide monomers to form a cross-linked gel network. The pH of the gel solution does not affect the deprotonation of TEMED or its ability to initiate polymerization. Therefore, the statement "At low pH, TEMED may become deprotonated and results in slower polymerization" is false.Rate this question:
-
- 38.
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.
A protein with MW of 40,000 daltons
- B.
A protein with MW of 23000 daltons
- C.
A protein with MW of 2,000 daltons
- D.
All of them will be in one band
Correct Answer
C. A protein with MW of 2,000 daltonsExplanation
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.Rate this question:
-
- 39.
The rate of migration of a protein in electrophoresis depends upon the
- A.
Time of electrophoresis
- B.
Electrical potential
- C.
Net charge on the protein
- D.
A, b, and c
- E.
B and c
Correct Answer
E. B and cExplanation
In electrophoresis, the rate of migration of a protein is influenced by the electrical potential applied and the net charge on the protein. The electrical potential determines the strength of the electric field, which affects the movement of charged particles like proteins. The net charge on the protein determines its overall charge, which can either attract or repel it towards the oppositely charged electrode. Therefore, both the electrical potential and the net charge on the protein play a role in determining its rate of migration in electrophoresis.Rate this question:
-
- 40.
A stain that does not interact with proteins in SDS-PAGE staining process and refrred to as negative stain. Which of the following stain show this characteristic? it is
- A.
Silver stain
- B.
Copper stain
- C.
Coomassie blue stain
- D.
None
Correct Answer
B. Copper stainExplanation
Copper stain is the correct answer because it does not interact with proteins in the SDS-PAGE staining process. Copper staining is commonly used to visualize nucleic acids, but it does not bind to proteins. Silver stain, on the other hand, is commonly used to detect proteins and other biomolecules. Coomassie blue stain is also used to stain proteins in SDS-PAGE, so it does interact with proteins. The option "none" is not a valid answer as it does not specify any stain.Rate this question:
-
- 41.
SDS-Page provides information on all of the following EXCEPT
- A.
Charge on subunit
- B.
Number and size of subunits
- C.
Purity
- D.
Molecular weight
Correct Answer
A. Charge on subunitExplanation
SDS-Page, or Sodium Dodecyl Sulfate-Polyacrylamide Gel Electrophoresis, is a technique used to separate proteins based on their size. It denatures the proteins and coats them with a negative charge, allowing them to migrate through the gel towards the positive electrode. Therefore, SDS-Page provides information on the number and size of subunits, purity, and molecular weight of proteins. However, it does not provide information on the charge on subunits, as the proteins are uniformly charged due to the SDS treatment.Rate this question:
-
- 42.
Smaller proteins will have a ______Rf value while larger proteins will have a _____Rf value during gel electrophoresis
- A.
Smaller, larger
- B.
Cannot be estimated
- C.
Larger, smaller
- D.
None of the above
Correct Answer
C. Larger, smallerExplanation
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.Rate this question:
-
- 43.
Assume that you are separating a protein with MW of approximately 45,000Da along with MW standard ranging from 6,500 to 66,200 Da using SDS-PAGE. Which percent of gel would you use for this sample?
- A.
4%
- B.
25%
- C.
12%
- D.
7.5%
Correct Answer
C. 12%Explanation
A protein with a molecular weight of approximately 45,000 Da would require a higher percentage of gel to separate it effectively using SDS-PAGE. Higher percentage gels provide better resolution for smaller proteins. Therefore, the correct answer of 12% is appropriate as it would allow for optimal separation of the protein of interest from the MW standard ranging from 6,500 to 66,200 Da.Rate this question:
-
- 44.
Polyacrylamide gels are prepared by polymerization of acrylamide monomer and N-N'-methylene-bis-acrylamide cross-linker in the presence of:
- A.
TEMED + b- mercaptoethaol
- B.
APS + b- mercaptoethaol
- C.
APS + TEMED
- D.
Tris- buffer + phosphate ion
Correct Answer
C. APS + TEMEDExplanation
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.Rate this question:
-
- 45.
Which type of gel separate proteins based on charge: mass ratio?
- A.
Native gel
- B.
2-D electrophoresis
- C.
Centrifugation
- D.
SDS-PAGE
Correct Answer
A. Native gelExplanation
A native gel is a type of gel that separates proteins based on their charge-to-mass ratio. It does not denature the proteins or disrupt their native structure, allowing them to maintain their natural charge. This enables the proteins to be separated solely based on their charge-to-mass ratio. In contrast, SDS-PAGE denatures proteins and separates them solely based on their mass. 2-D electrophoresis combines both charge and mass separation techniques to provide a more comprehensive separation of proteins. Centrifugation is a technique used for separating particles based on their size and density, not their charge-to-mass ratio.Rate this question:
-
- 46.
SDS binds to hydrophobic regions of the denatured protein chain in a constant ratio of:
- A.
2g SDS per gram protein
- B.
1 g SDS per gram protein
- C.
1 g SDS per 1.4 g protein
- D.
1.4 g SDS per gram protein
Correct Answer
D. 1.4 g SDS per gram proteinExplanation
SDS (sodium dodecyl sulfate) is a detergent commonly used in protein denaturation and electrophoresis. It binds to hydrophobic regions of the denatured protein chain. The correct answer, 1.4 g SDS per gram protein, suggests that there is a constant ratio between the amount of SDS and the amount of protein. For every gram of protein, 1.4 grams of SDS will bind to it. This ratio is important in experiments where SDS is used to denature proteins and separate them based on their molecular weight.Rate this question:
-
- 47.
In Western blot analysis, you used TMB (3'3'5,5'-Tetramethyl Benzidine) as a last step for visualizing unkonwn protein (ovalbumin). What is TMB?
- A.
A chromogenic substrate
- B.
An enzyme
- C.
Secondary antibody
- D.
Primary antibody
Correct Answer
A. A chromogenic substrateExplanation
TMB (3'3'5,5'-Tetramethyl Benzidine) is a chromogenic substrate used in Western blot analysis. It is added as the last step to visualize the unknown protein, ovalbumin. When TMB reacts with an enzyme, such as horseradish peroxidase (HRP), it produces a colored product that can be detected and quantified. Therefore, TMB acts as a chromogenic substrate, providing a visual signal for the presence of the protein of interest.Rate this question:
-
- 48.
Proteins that are separted in SDS-PAGE are subjected to treat with all of the following reagents/factor EXCEPT:
- A.
Tris-buffer
- B.
Ammonim (APS)
- C.
B-Mercatoethanol
- D.
SDS
- E.
Heat
Correct Answer
B. 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.Rate this question:
-
- 49.
Which of the following statement is NOT true for indirect method of protein detection of Western blot?
- A.
A wide variety of labeled secondary antibodies are available commercially
- B.
Immunoreactivity of primary antibody may be reduced as result of labeling
- C.
Same labeled secondary antibody can be used for detection, it is versatile
- D.
Immunoreactivity of the primary antibody is not affected by labeling
- E.
Different visualization markers can be used with the same primary antibody
Correct Answer
B. Immunoreactivity of primary antibody may be reduced as result of labelingExplanation
The immunoreactivity of the primary antibody may be reduced as a result of labeling.Rate this question:
-
- 50.
Cholesterol can easily be metabilized to CO2 and H2O in humans and excreted
- A.
False
- B.
True
Correct Answer
A. FalseExplanation
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.Rate this question:
-
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Feb 06, 2024Quiz Edited by
ProProfs Editorial Team -
Jul 22, 2011Quiz Created by
Ekanye
- Aeronautics Quizzes
- Aerospace Quizzes
- Agricultural Science Quizzes
- Astrology Quizzes
- Astronomy Quizzes
- Atom Quizzes
- Biochemistry Quizzes
- Biology Quizzes
- Biomechanics Quizzes
- Biostatistics Quizzes
- Biotechnology Quizzes
- Botany Quizzes
- Branches Of Science Quizzes
- Cytology Quizzes
- Easy Science Quizzes
- Ecology Quizzes
- Electrical Quizzes
- Embryology Quizzes
- Endocrinology Quizzes
- Engineering Quizzes
- Environmental Science Quizzes
- Epidemiology Quizzes
- Experiment Quizzes
- Forestry Quizzes
- Fossil Quizzes
- Gas Quizzes
- General Science Quizzes
- Genetics Quizzes
- Histology Quizzes
- Human Biology Quizzes
- Integrated Science Quizzes
- Invention Quizzes
- Library Science Quizzes
- Lighting Quizzes
- Liquid Quizzes
- Marine Biology Quizzes
- Microbiology Quizzes
- Molecular Biology Quizzes
- Nature Quizzes
- Neuroscience Quizzes
- Nuclear Science Quizzes
- Oceanography Quizzes
- Physics Quizzes
- Psychology Quizzes
- Science And Technology Quizzes
- Science Glossary Quizzes
- Science Knowledge Quizzes
- Science Practice Quizzes
- Scientific Method Quizzes
- Scientific Notation Quizzes
- Soil Science Quizzes
- Solar System Quizzes
- Solid Quizzes
- Toxicology Quizzes
- Zoology Quizzes