Help with Identification Techniques - GRE Subject Test: Biochemistry, Cell, and Molecular Biology

One function of DNA footprinting involves looking for binding sites by using nucleases to cleave DNA. Wherever the transcription factor is bound should be blocked and protected from cleavage. A sample of DNA can be run in a gel without transcription factors bound and compared to a gel run with DNA that has been exposed to the transcription factor. The difference in DNA sequence can be analyzed based on cleavage differences to identify the transcription factor binding region.

Northern blotting helps identify gene expression by looking at RNA. Western blotting helps identify specific proteins in a sample. PCR helps amplify a sample of DNA.

The secondary antibody is designed to interact specifically with the primary antibody. This is a crucial step because the secondary antibody will contain a detectable marker (usually a fluorescent marker).

The primary antibody interacts with the protein of interest. The membrane is blocked with milk or some other substance prior to addition of the antibodies to prevent nonspecific binding. Ponceau S dye is used to quickly and reversibly detect quantities of proteins on nitrocellulose membranes.

The correct answer is the student performed a gram stain and the bacteria is virulent. A gram stain specifically stains peptidoglycan, which is in the outer membrane layer of some bacteria, purple. The presence of peptidoglycan marks a bacteria as gram-positive. If peptidoglycan is not present, the bacteria will not stain purple, and the bacteria is said to be gram-negative. Gram-negative bacteria contain an outer layer of lipopolysaccharides that confer antibiotic resistance and virulence. Since we know that the bacteria in question doe not have an outer peptidoglycan layer, we know that it is virulent.

Northern blots are used to study RNA. When run on the gel used in a Northern, you can observe the length of the RNA (by how far it moves on the gel), the number of transcripts (the number of dark bands), and the expression level (how dark the bands are). Southern blots are used to study DNA, and Western blots are used to study proteins.

The correct answer is to determine which plasmids promote proliferative bacteria growth. In this experiment, the researcher wants to identify enzymes that, when over expressed in bacteria, allow the bacteria to grow in unfavorable (minimal) conditions. Since this researcher grows all 3000 transformed bacteria, each expressing a different enzyme, in one minimal media culture, he needs to be able to identify which enzymes promote growth. A barcode is an engineered unique DNA sequence that can be inserted on the plasmid expressing the enzyme. The researcher barcodes each unique plasmid. If a certain enzyme promotes bacteria growth, there will be bacterial replication and replication of the plasmid. Sequencing of the entire bacterial culture will determine which barcodes are overrepresented. Each barcode is then associated with the unique enzymes that allow for proliferative growth in minimal media.

The correct answer is co-immunoprecipitation. During this experiment, protein-protein complexes are cross linked then pulled down by an antibody specific to one of the proteins of interest. Then, these protein elutes are run on a western blot and probed for with a second antibody specific for the other protein believed to be in complex. An electrophoretic mobility shift assay measures the DNA-binding ability of a protein. A TUNEL assay is a measure for apoptosis by quantifying fragmented DNA. Protein binding microarrays determine the preferred DNA binding sequences for a given protein. ChIA-PET measures distal chromatin interactions within the genome.

The correct answer is none of the other answers. To increase specificity of the generated antibody, it is beneficial to expose the animal (in which the antibody will be made) to the entire purified protein, rather than a short synthesized predictive peptide. Moreover, affinity purifying the antibody by exposing the bled antisera to the purified protein and separated the bound antibody-protein fraction from antibody and protein alone will increase specificity. Following antibody generation, determining the optimal dilution of the antibody as well as appropriate blocking conditions to compete out non-specific binding with the specimen are necessary.

The correct answer is sectioning is preferred if the whole-mount is too large in the Z plane to resolve by confocal microscopy. Sectioning takes a small slice of an otherwise large specimen, and then subsequent antibody staining can be performed. Typically, this is preferable when one desires to stain certain cell types/layers or the entire specimen is too thick (large Z plane) to mount on a coverslip and resolve by microscopy.

The correct answer is the 1200 bp fragment will not image as brightly as the 3500 bp fragment. Ethidium bromide intercalates into DNA proportionally to the size of the fragment. Since we expect the same number of molecules of 1200 bp and 3500 bp fragments, the 1200 bp fragment should be roughly 3 times less bright than the 3500 bp fragment based on the number of ethidium bromide molecules incorporated into the strands. Additionally, smaller fragments migrate more quickly than larger fragments, and as such, are farther from the loading wells.

The correct answer is ribosomal RNA genes. These genes have highly conserved regions that scientists can use to design primers to amplify intervening regions. These intervening regions, however, are highly divergent and are extremely useful sequences to distinguish between species.

The resolution of an image is defined as the minimum distance between two points so that they can still be seen as two distinct structures. Resolution is important in microscopy. Images with clear resolution can be distinctly seen as two separate objects within the field of view.

One function of DNA footprinting involves looking for binding sites by using nucleases to cleave DNA. Wherever the transcription factor is bound should be blocked and protected from cleavage. A sample of DNA can be run in a gel without transcription factors bound and compared to a gel run with DNA that has been exposed to the transcription factor. The difference in DNA sequence can be analyzed based on cleavage differences to identify the transcription factor binding region.

Northern blotting helps identify gene expression by looking at RNA. Western blotting helps identify specific proteins in a sample. PCR helps amplify a sample of DNA.

The secondary antibody is designed to interact specifically with the primary antibody. This is a crucial step because the secondary antibody will contain a detectable marker (usually a fluorescent marker).

The primary antibody interacts with the protein of interest. The membrane is blocked with milk or some other substance prior to addition of the antibodies to prevent nonspecific binding. Ponceau S dye is used to quickly and reversibly detect quantities of proteins on nitrocellulose membranes.

The correct answer is the student performed a gram stain and the bacteria is virulent. A gram stain specifically stains peptidoglycan, which is in the outer membrane layer of some bacteria, purple. The presence of peptidoglycan marks a bacteria as gram-positive. If peptidoglycan is not present, the bacteria will not stain purple, and the bacteria is said to be gram-negative. Gram-negative bacteria contain an outer layer of lipopolysaccharides that confer antibiotic resistance and virulence. Since we know that the bacteria in question doe not have an outer peptidoglycan layer, we know that it is virulent.

Northern blots are used to study RNA. When run on the gel used in a Northern, you can observe the length of the RNA (by how far it moves on the gel), the number of transcripts (the number of dark bands), and the expression level (how dark the bands are). Southern blots are used to study DNA, and Western blots are used to study proteins.

The correct answer is to determine which plasmids promote proliferative bacteria growth. In this experiment, the researcher wants to identify enzymes that, when over expressed in bacteria, allow the bacteria to grow in unfavorable (minimal) conditions. Since this researcher grows all 3000 transformed bacteria, each expressing a different enzyme, in one minimal media culture, he needs to be able to identify which enzymes promote growth. A barcode is an engineered unique DNA sequence that can be inserted on the plasmid expressing the enzyme. The researcher barcodes each unique plasmid. If a certain enzyme promotes bacteria growth, there will be bacterial replication and replication of the plasmid. Sequencing of the entire bacterial culture will determine which barcodes are overrepresented. Each barcode is then associated with the unique enzymes that allow for proliferative growth in minimal media.

The correct answer is co-immunoprecipitation. During this experiment, protein-protein complexes are cross linked then pulled down by an antibody specific to one of the proteins of interest. Then, these protein elutes are run on a western blot and probed for with a second antibody specific for the other protein believed to be in complex. An electrophoretic mobility shift assay measures the DNA-binding ability of a protein. A TUNEL assay is a measure for apoptosis by quantifying fragmented DNA. Protein binding microarrays determine the preferred DNA binding sequences for a given protein. ChIA-PET measures distal chromatin interactions within the genome.

The correct answer is none of the other answers. To increase specificity of the generated antibody, it is beneficial to expose the animal (in which the antibody will be made) to the entire purified protein, rather than a short synthesized predictive peptide. Moreover, affinity purifying the antibody by exposing the bled antisera to the purified protein and separated the bound antibody-protein fraction from antibody and protein alone will increase specificity. Following antibody generation, determining the optimal dilution of the antibody as well as appropriate blocking conditions to compete out non-specific binding with the specimen are necessary.

The correct answer is sectioning is preferred if the whole-mount is too large in the Z plane to resolve by confocal microscopy. Sectioning takes a small slice of an otherwise large specimen, and then subsequent antibody staining can be performed. Typically, this is preferable when one desires to stain certain cell types/layers or the entire specimen is too thick (large Z plane) to mount on a coverslip and resolve by microscopy.

The correct answer is the 1200 bp fragment will not image as brightly as the 3500 bp fragment. Ethidium bromide intercalates into DNA proportionally to the size of the fragment. Since we expect the same number of molecules of 1200 bp and 3500 bp fragments, the 1200 bp fragment should be roughly 3 times less bright than the 3500 bp fragment based on the number of ethidium bromide molecules incorporated into the strands. Additionally, smaller fragments migrate more quickly than larger fragments, and as such, are farther from the loading wells.