GRE Subject Test: Biology : Lab Techniques
Study concepts, example questions & explanations for GRE Subject Test: Biology
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Example Questions
Example Question #61 : Lab Techniques
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?
Electrophoretic mobility shift assay (ESMA)
Mass spectrophotometry
Immunohistochemistry
Western blot
Electroporation
Mass spectrophotometry
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.
Example Question #62 : Lab Techniques
A student researcher wants to determine if two proteins physically interact within a cell. What technique is best suitable for this experiment?
Cross-linking
Chromatin immunoprecipitation sequencing
Chromatin interaction analysis by paired-end tag sequencing
Immunoprecipitation
Co-immunoprecipitation
Co-immunoprecipitation
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.
Example Question #11 : Other Lab Techniques
The process by which anatomical specimens are separated and analyzed in a detailed manner is termed __________.
dissociation
dissemination
evisceration
dissection
exsanguination
dissection
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.
Example Question #2 : Understanding Other Techniques
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?
Pre-incubation 30min, heat shock duration 60sec, heat shock temp 39°C
Pre-incubation 60min, heat shock duration 30sec, heat shock temp 42°C
Pre-incubation 30min, heat shock duration 60sec, heat shock temp 42°C
Pre-incubation 60min, heat shock duration 60sec, heat shock temp 42°C
Pre-incubation 30min, heat shock duration 30sec, heat shock temp 42°C
Pre-incubation 60min, heat shock duration 60sec, heat shock temp 42°C
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.
Example Question #2 : Understanding Other Techniques
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?
Heat shock temp
Pre-incubation duration
Transformation efficiency
Heat shock duration
A combinatorial effect of the variables tested
Heat shock temp
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.
Example Question #4 : Understanding Other Techniques
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.
Example Question #3 : Understanding Other Techniques
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?
Conditions 4, 5, and 6
Conditions 1, 2, and 3
Conditions 5, 8, and 9
Conditions 4, 6, and 8
Conditions 7, 8, and 9
Conditions 4, 5, and 6
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.
Example Question #6 : Understanding Other Techniques
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?
Water incubators verses air-vented incubators
None of the other answers
Varying concentrations of ampicillin
Chemical competent bacteria verses electro-competent bacteria
Varying amounts of plasmid DNA
Varying amounts of plasmid DNA
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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