Other Lab Techniques - GRE Subject Test: Biology
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Which of the following stains could be used to identify bacteria?
Which of the following stains could be used to identify bacteria?
Gram staining is used to distinguish different types of bacteria in order to aid in their identification. Since bacteria are generally transparent, they must be stained before viewing under the microscope. Gram-positive microorganisms stain purple and Gram-negative microorganisms stain red or pink.
Acid fast stain is used to identify the causative agent of tuberculosis. White blood cells are differentiated after being stained with Wright's stain. Giemsa is used to stain blood cells and chromosomes. Tissues are stained with eosin.
Gram staining is used to distinguish different types of bacteria in order to aid in their identification. Since bacteria are generally transparent, they must be stained before viewing under the microscope. Gram-positive microorganisms stain purple and Gram-negative microorganisms stain red or pink.
Acid fast stain is used to identify the causative agent of tuberculosis. White blood cells are differentiated after being stained with Wright's stain. Giemsa is used to stain blood cells and chromosomes. Tissues are stained with eosin.
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A student co-immunoprecipitates the NF-kappa B transcription factor that is bound to a protein, which the student cannot identify. What technique would be suitable for identifying this unknown protein?
A student co-immunoprecipitates the NF-kappa B transcription factor that is bound to a protein, which the student cannot identify. What technique would be suitable for identifying this unknown protein?
The correct answer is mass spectrophotometry. Mass spec uses ions to measure the mass-to-charge ratio of the given molecule. This signature ratio can then be compared to other known mass-to-charge ratios of other proteins, allowing for identification of the unknown protein based on similar ratios.
The correct answer is mass spectrophotometry. Mass spec uses ions to measure the mass-to-charge ratio of the given molecule. This signature ratio can then be compared to other known mass-to-charge ratios of other proteins, allowing for identification of the unknown protein based on similar ratios.
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A student researcher wants to determine if two proteins physically interact within a cell. What technique is best suitable for this experiment?
A student researcher wants to determine if two proteins physically interact within a cell. What technique is best suitable for this experiment?
The correct answer is co-immunoprecipitation. Protein-protein interactions are cross-linked within a cell. Using an antibody to the first protein, this protein (along with any molecules bound to it) is isolated from the cell lysate. Then using a second antibody to the second protein of interest, a western blot is performed to determine if the second antibody was pulled down in the first immunoprecipitation with antibody number 1. Immunoprecipitation involves isolating a particular protein out of a solution. Chromatin immunoprecipitation sequencing involves analysis of interactions between DNA and proteins, and is used to identify the binding sites of proteins associated with DNA. Cross linking is the process by which two polymers are joined, it does not indicate whether two proteins were interacting prior to the procedure. Chromatin interaction analysis by paired-end tag sequencing determines chromatin interaction, not protein interaction.
The correct answer is co-immunoprecipitation. Protein-protein interactions are cross-linked within a cell. Using an antibody to the first protein, this protein (along with any molecules bound to it) is isolated from the cell lysate. Then using a second antibody to the second protein of interest, a western blot is performed to determine if the second antibody was pulled down in the first immunoprecipitation with antibody number 1. Immunoprecipitation involves isolating a particular protein out of a solution. Chromatin immunoprecipitation sequencing involves analysis of interactions between DNA and proteins, and is used to identify the binding sites of proteins associated with DNA. Cross linking is the process by which two polymers are joined, it does not indicate whether two proteins were interacting prior to the procedure. Chromatin interaction analysis by paired-end tag sequencing determines chromatin interaction, not protein interaction.
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The process by which anatomical specimens are separated and analyzed in a detailed manner is termed __________.
The process by which anatomical specimens are separated and analyzed in a detailed manner is termed __________.
Dissection is a surgical procedure in which instruments are used to cut and separate tissues for study.
Eviseration is the removal of the organs or the contents of a cavity. Exsanguination describes massive bleeding. Dissociation is the separation of a complex compound into simpler molecules. Dissemination involves spreading something over a considerable area.
Dissection is a surgical procedure in which instruments are used to cut and separate tissues for study.
Eviseration is the removal of the organs or the contents of a cavity. Exsanguination describes massive bleeding. Dissociation is the separation of a complex compound into simpler molecules. Dissemination involves spreading something over a considerable area.
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Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Of the conditions tested, which condition is optimal to generate a high transformation yield?
Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Of the conditions tested, which condition is optimal to generate a high transformation yield?
High transformation yield corresponds with the highest transformation efficiency. The condition that had a pre-incubation of 60 minutes, a heat shock duration of 60 seconds, and a heat shock temperature of 42°C had the highest transformation efficiency of 5.5 x 107 cfu/μg.
High transformation yield corresponds with the highest transformation efficiency. The condition that had a pre-incubation of 60 minutes, a heat shock duration of 60 seconds, and a heat shock temperature of 42°C had the highest transformation efficiency of 5.5 x 107 cfu/μg.
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Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Which variable had the greatest effect on a high transformation yield?
Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Which variable had the greatest effect on a high transformation yield?
Differences in pre-incubation duration and heat shock duration do not have a large effect on transformation efficiency. However, at a heat shock temperature of 42°C degrees, regardless of the pre-incubation and heat shock durations, the transformation efficiency is at least 105 fold bigger. Thus, heat shock temperature is the most potent variable tested.
Differences in pre-incubation duration and heat shock duration do not have a large effect on transformation efficiency. However, at a heat shock temperature of 42°C degrees, regardless of the pre-incubation and heat shock durations, the transformation efficiency is at least 105 fold bigger. Thus, heat shock temperature is the most potent variable tested.
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Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
If the student researcher had also done trials with a heat shock temperature of 36°C, the resulting transformation yield would most likely be __________.
Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
If the student researcher had also done trials with a heat shock temperature of 36°C, the resulting transformation yield would most likely be __________.
For a successful transformation, the heat shock temperature must be higher than the temperature at which the bacteria grows (37°C). This is also the temperature at which the bacteria recover. The heat shock creates pores in the plasma membrane of the cell, allowing the plasmid DNA to enter. If the temperature is not high enough, pores will not form in the membrane and the bacteria will not be transformed; therefore, the transformation efficiency will be zero.
For a successful transformation, the heat shock temperature must be higher than the temperature at which the bacteria grows (37°C). This is also the temperature at which the bacteria recover. The heat shock creates pores in the plasma membrane of the cell, allowing the plasmid DNA to enter. If the temperature is not high enough, pores will not form in the membrane and the bacteria will not be transformed; therefore, the transformation efficiency will be zero.
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Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
If the student realizes that a fraction of the ampicillin agar plates he used to select for transformed bacteria had expired, which conditions would most likely have been grown on expired ampicillin?
Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
If the student realizes that a fraction of the ampicillin agar plates he used to select for transformed bacteria had expired, which conditions would most likely have been grown on expired ampicillin?
If the ampicillin expired, there is no selection for only transformed bacteria to grow. As a result, both transformed and non-transformed bacteria would grow. If the ampicillin were indeed expired, we would expect many more colonies to grow. Conditions 4, 5, and 6 have the highest transformation efficiency (most colonies) and would have most likely been the conditions plated on expired ampicillin.
If the ampicillin expired, there is no selection for only transformed bacteria to grow. As a result, both transformed and non-transformed bacteria would grow. If the ampicillin were indeed expired, we would expect many more colonies to grow. Conditions 4, 5, and 6 have the highest transformation efficiency (most colonies) and would have most likely been the conditions plated on expired ampicillin.
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Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Besides the indicated variables in table 1, what is most likely the next variable the student would introduce in this experiment to optimize transformation efficiency?
Bacterial transformations are a useful tool for biologists to replicate valuable plasmid DNA so that it can be used for a myriad of experiments. A student researcher has designed an experiment to determine the effects of pre-incubation duration, heat shock duration, and heat shock temperature on transformation efficiency. The student transformed 1μg of a β**-**lactamase encoding plasmid and recovered the bacteria for 1 hour at 37°C (shaking at 225rpm) for all conditions in Table 1. He then plated the transformed bacteria on ampicillin agar places and observed growth 24 hours later. Table 1 outlines the variables in the experiment.
Besides the indicated variables in table 1, what is most likely the next variable the student would introduce in this experiment to optimize transformation efficiency?
The correct answer is varying amounts of DNA. Increasing the amount of plasmid DNA can increase transformation efficiency; however, if too much plasmid DNA is introduced into the transformation, a lawn of bacteria will result.
Electro-competent bacteria are transformed by an entirely different mechanism than described here. The type of incubator used for the heat shock will most likely not produce a noticeable difference. When troubleshooting a transformation, the amount of ampicillin is rarely considered; the amount of ampicillin used to select for transformed bacteria is generally always the same.
The correct answer is varying amounts of DNA. Increasing the amount of plasmid DNA can increase transformation efficiency; however, if too much plasmid DNA is introduced into the transformation, a lawn of bacteria will result.
Electro-competent bacteria are transformed by an entirely different mechanism than described here. The type of incubator used for the heat shock will most likely not produce a noticeable difference. When troubleshooting a transformation, the amount of ampicillin is rarely considered; the amount of ampicillin used to select for transformed bacteria is generally always the same.
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When would it be appropriate to use extraction in order to separate compounds in a solution?
When would it be appropriate to use extraction in order to separate compounds in a solution?
Extraction is a useful separation technique when there is a mixture of compounds in a solution that have similar polarities, but different solubilities. The three-step process of extraction can take advantage of different solubilities by introducing the mixture to different acidic and basic conditions.
Extraction is a useful separation technique when there is a mixture of compounds in a solution that have similar polarities, but different solubilities. The three-step process of extraction can take advantage of different solubilities by introducing the mixture to different acidic and basic conditions.
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In polymerase chain reaction (PCR), the reaction mixture is heated to approximately 98°C during the first cycling step in the procedure. Which of the following describes the purpose of this step?
In polymerase chain reaction (PCR), the reaction mixture is heated to approximately 98°C during the first cycling step in the procedure. Which of the following describes the purpose of this step?
Heating of the reaction to roughly 98°C is required to separate the template DNA strands from one another, thereby producing single strands that can pair to their complementary primer. At this temperature, the hydrogen bonds between base pairs break and the strands separate.
Taq polymerase doesn't require this temperature to be active, non-target sequences are not degraded during PCR, the dNTPs are already soluble, and the optimal annealing temperature for primers is actually much lower than 98°C.
Heating of the reaction to roughly 98°C is required to separate the template DNA strands from one another, thereby producing single strands that can pair to their complementary primer. At this temperature, the hydrogen bonds between base pairs break and the strands separate.
Taq polymerase doesn't require this temperature to be active, non-target sequences are not degraded during PCR, the dNTPs are already soluble, and the optimal annealing temperature for primers is actually much lower than 98°C.
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Which technique is best suited to determine if a protein is active and able to bind DNA?
Which technique is best suited to determine if a protein is active and able to bind DNA?
In an EMSA, a radiolabeled sequence of DNA that the protein of interest normally binds is incubated with the protein. This mixutre is then run on a non-denaturing gel. If the protein binds the radiolabeled DNA sequence, radioacitivity will be detected towards the top of the gel; however, if it does not bind, the radiolabeled DNA probe alone will run more quickly towards the bottom of the gel.
In an EMSA, a radiolabeled sequence of DNA that the protein of interest normally binds is incubated with the protein. This mixutre is then run on a non-denaturing gel. If the protein binds the radiolabeled DNA sequence, radioacitivity will be detected towards the top of the gel; however, if it does not bind, the radiolabeled DNA probe alone will run more quickly towards the bottom of the gel.
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What is the purpose of a electrophoretic mobility supershift assay?
What is the purpose of a electrophoretic mobility supershift assay?
The purpose of an electrophoretic mobility supershift assay (EMSA) is to determine if a certain protein binds a DNA sequence. EMSAs determine if a protein is "active", meaning that it is capable of binding its target DNA sequence. In an EMSA, a radiolabelled DNA fragment is incubated with cellular protein lysates, then run on a non-denaturing gel. Any protein-bound DNA fragments will migrate slower than unbound DNA fragments. When performing a supershift, one wants to determine if a specific protein binds a radiolabelled DNA probe by use of an antibody. If the antibody binds the protein-DNA complex, this will migrate even slower than the protein-DNA complex alone.
The purpose of an electrophoretic mobility supershift assay (EMSA) is to determine if a certain protein binds a DNA sequence. EMSAs determine if a protein is "active", meaning that it is capable of binding its target DNA sequence. In an EMSA, a radiolabelled DNA fragment is incubated with cellular protein lysates, then run on a non-denaturing gel. Any protein-bound DNA fragments will migrate slower than unbound DNA fragments. When performing a supershift, one wants to determine if a specific protein binds a radiolabelled DNA probe by use of an antibody. If the antibody binds the protein-DNA complex, this will migrate even slower than the protein-DNA complex alone.
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Capillary electrophoresis instrumentation involves which of the following components?
Capillary electrophoresis instrumentation involves which of the following components?
A buffer system is required in a capillary electrophoresis system in order to supply the ions necessary to carry the electric current. Ions become depleted quickly and the buffer system will need to be replenished regularly.
The positive and negative electrodes, in combination with a high voltage power supply create the current. The source of the sample and destination of the sample will have opposite charged electrodes. Migration of the analytes through the polymer filled capillary will depend on the applied electric field and their size.
The detector of a capillary electrophoresis system will vary based on the instrumentation type. It is used to detect the size of the molecules and the speed at which they move through the matrix. Sizing methods and databases are then used for analysis.
Polymer serves as the separation matrix of the system. The viscosity of the polymer in combination with the electroosmotic flow of they system will separate molecules based on their size.
A buffer system is required in a capillary electrophoresis system in order to supply the ions necessary to carry the electric current. Ions become depleted quickly and the buffer system will need to be replenished regularly.
The positive and negative electrodes, in combination with a high voltage power supply create the current. The source of the sample and destination of the sample will have opposite charged electrodes. Migration of the analytes through the polymer filled capillary will depend on the applied electric field and their size.
The detector of a capillary electrophoresis system will vary based on the instrumentation type. It is used to detect the size of the molecules and the speed at which they move through the matrix. Sizing methods and databases are then used for analysis.
Polymer serves as the separation matrix of the system. The viscosity of the polymer in combination with the electroosmotic flow of they system will separate molecules based on their size.
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Compared to slab gel electrophoresis, which is an advantage of capillary electrophoresis?
Compared to slab gel electrophoresis, which is an advantage of capillary electrophoresis?
Capillary electrophoresis allows for better reproducibility of liquid polymer throughout a capillary compared to slab gels. Much of the time gels are manually poured and uneven gel thickness can occur.
Real time data viewing is possible on a capillary electrophoresis instrument's system computer.
Higher sample consumption is common with slab gels. More sample is required to be loaded in each lane. If retesting is necessary, the sample must be prepared and loaded into a new gel. Very small quantities of sample are consumed in the injection step into the capillary. Samples can be easily retested through reinjection from the original sample vial.
Capillary electrophoresis has the possibility of being a completely automated process, including automated sample loading. This saves analyst time during sample prep, injection, and separation. Gels require manual sample loading and some instruments require gel handling for scanning or photography after the electrophoresis process.
Capillary electrophoresis allows for better reproducibility of liquid polymer throughout a capillary compared to slab gels. Much of the time gels are manually poured and uneven gel thickness can occur.
Real time data viewing is possible on a capillary electrophoresis instrument's system computer.
Higher sample consumption is common with slab gels. More sample is required to be loaded in each lane. If retesting is necessary, the sample must be prepared and loaded into a new gel. Very small quantities of sample are consumed in the injection step into the capillary. Samples can be easily retested through reinjection from the original sample vial.
Capillary electrophoresis has the possibility of being a completely automated process, including automated sample loading. This saves analyst time during sample prep, injection, and separation. Gels require manual sample loading and some instruments require gel handling for scanning or photography after the electrophoresis process.
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Which of the following is a method for sample introduction into a capillary electrophoresis system?
Which of the following is a method for sample introduction into a capillary electrophoresis system?
All of these methods require the immersion of the capillary end into the sample for introduction into the capillary electrophoresis system.
All of these methods require the immersion of the capillary end into the sample for introduction into the capillary electrophoresis system.
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Which of the following is a method for sample introduction into a capillary electrophoresis system?
Which of the following is a method for sample introduction into a capillary electrophoresis system?
All of these methods require the immersion of the capillary end into the sample for introduction into the capillary electrophoresis system.
All of these methods require the immersion of the capillary end into the sample for introduction into the capillary electrophoresis system.
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Which of the following stains could be used to identify bacteria?
Which of the following stains could be used to identify bacteria?
Gram staining is used to distinguish different types of bacteria in order to aid in their identification. Since bacteria are generally transparent, they must be stained before viewing under the microscope. Gram-positive microorganisms stain purple and Gram-negative microorganisms stain red or pink.
Acid fast stain is used to identify the causative agent of tuberculosis. White blood cells are differentiated after being stained with Wright's stain. Giemsa is used to stain blood cells and chromosomes. Tissues are stained with eosin.
Gram staining is used to distinguish different types of bacteria in order to aid in their identification. Since bacteria are generally transparent, they must be stained before viewing under the microscope. Gram-positive microorganisms stain purple and Gram-negative microorganisms stain red or pink.
Acid fast stain is used to identify the causative agent of tuberculosis. White blood cells are differentiated after being stained with Wright's stain. Giemsa is used to stain blood cells and chromosomes. Tissues are stained with eosin.
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A student co-immunoprecipitates the NF-kappa B transcription factor that is bound to a protein, which the student cannot identify. What technique would be suitable for identifying this unknown protein?
A student co-immunoprecipitates the NF-kappa B transcription factor that is bound to a protein, which the student cannot identify. What technique would be suitable for identifying this unknown protein?
The correct answer is mass spectrophotometry. Mass spec uses ions to measure the mass-to-charge ratio of the given molecule. This signature ratio can then be compared to other known mass-to-charge ratios of other proteins, allowing for identification of the unknown protein based on similar ratios.
The correct answer is mass spectrophotometry. Mass spec uses ions to measure the mass-to-charge ratio of the given molecule. This signature ratio can then be compared to other known mass-to-charge ratios of other proteins, allowing for identification of the unknown protein based on similar ratios.
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A student researcher wants to determine if two proteins physically interact within a cell. What technique is best suitable for this experiment?
A student researcher wants to determine if two proteins physically interact within a cell. What technique is best suitable for this experiment?
The correct answer is co-immunoprecipitation. Protein-protein interactions are cross-linked within a cell. Using an antibody to the first protein, this protein (along with any molecules bound to it) is isolated from the cell lysate. Then using a second antibody to the second protein of interest, a western blot is performed to determine if the second antibody was pulled down in the first immunoprecipitation with antibody number 1. Immunoprecipitation involves isolating a particular protein out of a solution. Chromatin immunoprecipitation sequencing involves analysis of interactions between DNA and proteins, and is used to identify the binding sites of proteins associated with DNA. Cross linking is the process by which two polymers are joined, it does not indicate whether two proteins were interacting prior to the procedure. Chromatin interaction analysis by paired-end tag sequencing determines chromatin interaction, not protein interaction.
The correct answer is co-immunoprecipitation. Protein-protein interactions are cross-linked within a cell. Using an antibody to the first protein, this protein (along with any molecules bound to it) is isolated from the cell lysate. Then using a second antibody to the second protein of interest, a western blot is performed to determine if the second antibody was pulled down in the first immunoprecipitation with antibody number 1. Immunoprecipitation involves isolating a particular protein out of a solution. Chromatin immunoprecipitation sequencing involves analysis of interactions between DNA and proteins, and is used to identify the binding sites of proteins associated with DNA. Cross linking is the process by which two polymers are joined, it does not indicate whether two proteins were interacting prior to the procedure. Chromatin interaction analysis by paired-end tag sequencing determines chromatin interaction, not protein interaction.
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