Deck 15: Genomics and Proteomics
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Deck 15: Genomics and Proteomics
1
Which technique would NOT be used to find a gene for a functional protein in a sequenced region of a genome?
A)Scan the region for ORFs.
B)See if there are restriction sites in the region.
C)See if an SNP database contains sequences in the region.
D)Scan the region for promoter sequences.
E)Scan the region for intron splice sites.
A)Scan the region for ORFs.
B)See if there are restriction sites in the region.
C)See if an SNP database contains sequences in the region.
D)Scan the region for promoter sequences.
E)Scan the region for intron splice sites.
C
2
A section of a genome is cut with three enzymes: A, B, and C.Cutting with A and B yields a 10-kb fragment.Cutting with B and C yields a 2-kb fragment.What is the expected result from a digest with A and C, if the C site lies between the A and B sites?
A)A 12-kb fragment
B)An 8-kb fragment
C)An 8-kb and a 2-kb fragment
D)A 10-kb and a 2-kb fragment
E)A 10-kb, an 8-kb, and a 2-kb fragment
A)A 12-kb fragment
B)An 8-kb fragment
C)An 8-kb and a 2-kb fragment
D)A 10-kb and a 2-kb fragment
E)A 10-kb, an 8-kb, and a 2-kb fragment
B
3
In the following list of cloned fragments, the fragments needed to make the longest possible contig with the least amount of overlap are: 
A)A, B, F.
B)A, B, C, E.
C)B, C, D, F.
D)B, C, D, E, F.
E)A, B, C, D, E, F.

A)A, B, F.
B)A, B, C, E.
C)B, C, D, F.
D)B, C, D, E, F.
E)A, B, C, D, E, F.
C
4
Which of the following is NOT a reason that metagenomics is particularly useful for studying microbes?
A)Many bacteria that cannot be cultured in the laboratory can be now studied.
B)The composition of natural microbial communities can be reconstructed.
C)The microbiome of individuals can be compared to understand the basis of disease.
D)Previously unknown species of bacteria and new gene families have been discovered.
E)The genomes of somatic cells in a human individual can vary significantly.
A)Many bacteria that cannot be cultured in the laboratory can be now studied.
B)The composition of natural microbial communities can be reconstructed.
C)The microbiome of individuals can be compared to understand the basis of disease.
D)Previously unknown species of bacteria and new gene families have been discovered.
E)The genomes of somatic cells in a human individual can vary significantly.
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5
Crossing over is often reduced around centrometric regions of chromosomes.If you were trying to construct a genetic map of two linked marker loci in this region, what result might you obtain?
A)The recombination frequencies would be low, and you would deduce that the markers were very close to one another.
B)The recombination frequencies would be low, and you would deduce that the markers were very far from one another.
C)The recombination frequencies would be high, and you would deduce that the markers were very close to one another.
D)The recombination frequencies would be high, and you would deduce that the markers were very far from one another.
E)The recombination frequencies would be too low to detect, and you would not be able to estimate the distance between genetic markers.
A)The recombination frequencies would be low, and you would deduce that the markers were very close to one another.
B)The recombination frequencies would be low, and you would deduce that the markers were very far from one another.
C)The recombination frequencies would be high, and you would deduce that the markers were very close to one another.
D)The recombination frequencies would be high, and you would deduce that the markers were very far from one another.
E)The recombination frequencies would be too low to detect, and you would not be able to estimate the distance between genetic markers.
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6
Loci that are far apart:
A)have a higher recombination rate than loci that are close together.
B)exhibit a recombination frequency of greater than 50%.
C)always segregate together in meiosis.
D)influence the phenotype to a lesser degree than loci that are close together.
E)have less attraction than close loci.
A)have a higher recombination rate than loci that are close together.
B)exhibit a recombination frequency of greater than 50%.
C)always segregate together in meiosis.
D)influence the phenotype to a lesser degree than loci that are close together.
E)have less attraction than close loci.
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7
For a physical map of a chromosome, distances are measured in units of:
A)SNPs.
B)RFLPs.
C)centiMorgans.
D)base pairs.
E)contigs.
A)SNPs.
B)RFLPs.
C)centiMorgans.
D)base pairs.
E)contigs.
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8
Which of the following is a disadvantage of map-based sequencing over shotgun sequencing?
A)Map-based sequencing minimizes the amount of repeat sequencing in which the same region is sequenced several times.
B)Map-based sequencing yields well-characterized physical maps as well as actual sequences.
C)The relationship of overlapping fragments to sequences is known before sequencing begins.
D)Map-based sequencing is more time-consuming because extensive mapping is required before sequencing starts.
E)Restriction patterns can be used to identify overlapping clones.
A)Map-based sequencing minimizes the amount of repeat sequencing in which the same region is sequenced several times.
B)Map-based sequencing yields well-characterized physical maps as well as actual sequences.
C)The relationship of overlapping fragments to sequences is known before sequencing begins.
D)Map-based sequencing is more time-consuming because extensive mapping is required before sequencing starts.
E)Restriction patterns can be used to identify overlapping clones.
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9
A human gene with a disease phenotype is going to be mapped by positional cloning.Which would be the MOST useful for this task?
A)Information about bacterial homologs of the gene
B)A restriction enzyme map of the human genome
C)Microarray data of tissues in which the gene is expressed
D)Data about the inheritance of SNP markers in families with the disease
E)Whole-genome-shotgun clones of the human genome
A)Information about bacterial homologs of the gene
B)A restriction enzyme map of the human genome
C)Microarray data of tissues in which the gene is expressed
D)Data about the inheritance of SNP markers in families with the disease
E)Whole-genome-shotgun clones of the human genome
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10
If a restriction enzyme cuts a circular DNA into three fragments, how many restriction sites are there in the DNA?
A)Two
B)Three
C)Four
D)Six
E)Five
A)Two
B)Three
C)Four
D)Six
E)Five
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11
A BLAST search is done to:
A)scan the database for a particular sequence.
B)find the chromosomal location of a sequence.
C)predict the three-dimensional structure of a protein from its amino acid sequence.
D)find restriction sites and SNPs in a sequence.
E)determine the conditions under which a gene is expressed.
A)scan the database for a particular sequence.
B)find the chromosomal location of a sequence.
C)predict the three-dimensional structure of a protein from its amino acid sequence.
D)find restriction sites and SNPs in a sequence.
E)determine the conditions under which a gene is expressed.
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12
You cloned a 2-kb fragment of DNA into a bacterial cloning vector and want to construct a restriction map of the insert.You amplify the 2-kb insert using PCR, purify it, and subject it to differential digestion with the enzymes EcoRI and HindIII, gel-fractionate the digests, and visualize the restriction patterns by staining the gels with ethidium bromide to generate the following results.Using these data, which two loci would be CLOSEST on a genetic map if the frequency of crossing over is equal in this region? 
A)The EcoRI sites that are 100 bp apart would show the smallest distance in cM.
B)The HindIII sites that are 750 bp apart would show the smallest distance in cM.
C)The HindIII site and the EcoRI site that are 200 bp apart would show the smallest distances in cM.
D)The EcoRI site and the HindIII site that are 300 bp apart would show the smallest distance in cM.
E)The EcoRI sites that are 150 bp apart would show the smallest distance in cM.

A)The EcoRI sites that are 100 bp apart would show the smallest distance in cM.
B)The HindIII sites that are 750 bp apart would show the smallest distance in cM.
C)The HindIII site and the EcoRI site that are 200 bp apart would show the smallest distances in cM.
D)The EcoRI site and the HindIII site that are 300 bp apart would show the smallest distance in cM.
E)The EcoRI sites that are 150 bp apart would show the smallest distance in cM.
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13
You cloned a 2-kb fragment of DNA into a bacterial cloning vector and want to construct a restriction map of the insert.You amplify the 2-kb insert using PCR, purify it, and subject it to differential digestion with the enzymes EcoRI and HindIII, gel-fractionate the digests, and visualize the restriction patterns by staining the gels with ethidium bromide to generate the following results.Using these data, indicate the CORRECT order of restriction sites in the DNA insert. 
A)EcoRI, HindIII, EcoRI, HindIII, EcoRI
B)HindIII, EcoRI, HindIII, EcoRI, HindIII
C)HindIII, EcoRI, HindIII, HindIII, EcoRI
D)HindIII, EcoRI, EcoRI, HindIII, EcoRI
E)EcoRI, EcoRI, HindIII, EcoRI, HindIII

A)EcoRI, HindIII, EcoRI, HindIII, EcoRI
B)HindIII, EcoRI, HindIII, EcoRI, HindIII
C)HindIII, EcoRI, HindIII, HindIII, EcoRI
D)HindIII, EcoRI, EcoRI, HindIII, EcoRI
E)EcoRI, EcoRI, HindIII, EcoRI, HindIII
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14
Which of the following is TRUE of genetic maps?
A)They are derived from direct analysis of DNA.
B)They are based on frequencies of recombination between loci.
C)They have a high level of detail rivaling that of physical maps.
D)They always accurately correspond to physical distances between genes.
E)They are more accurate than physical maps.
A)They are derived from direct analysis of DNA.
B)They are based on frequencies of recombination between loci.
C)They have a high level of detail rivaling that of physical maps.
D)They always accurately correspond to physical distances between genes.
E)They are more accurate than physical maps.
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15
Two unrelated individuals are 99.9% similar.Since the human genome consists of roughly 3 billion bases, these two individuals will have _____ bases that are different.
A)3 million
B)3 billion
C)1 million
D)300 million
E)This number cannot be calculated from these data.
A)3 million
B)3 billion
C)1 million
D)300 million
E)This number cannot be calculated from these data.
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16
For a genetic map of a chromosome, distances are measured in units of:
A)SNPs.
B)RFLPs.
C)centiMorgans.
D)base pairs.
E)contigs.
A)SNPs.
B)RFLPs.
C)centiMorgans.
D)base pairs.
E)contigs.
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17
How is the microbiome involved in obesity?
A)Lean people have fewer Firmicutes than obese people.
B)Obese people have fewer Firmicutes than lean people.
C)Obese people are able to extract more calories from their food because of their Firmicutes.
D)Lean people have fewer Firmicutes than obese people, and obese people are able to extract more calories from their food because of their Firmicutes.
E)Obese people have fewer Firmicutes than lean people, and they are able to extract more calories from their food because of their Firmicutes.
A)Lean people have fewer Firmicutes than obese people.
B)Obese people have fewer Firmicutes than lean people.
C)Obese people are able to extract more calories from their food because of their Firmicutes.
D)Lean people have fewer Firmicutes than obese people, and obese people are able to extract more calories from their food because of their Firmicutes.
E)Obese people have fewer Firmicutes than lean people, and they are able to extract more calories from their food because of their Firmicutes.
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18
A linear DNA fragment is cut with a restriction enzyme to yield two fragments.How many restriction sites are there for the enzyme in this fragment?
A)1
B)2
C)3
D)4
E)Cannot be determined with this information.
A)1
B)2
C)3
D)4
E)Cannot be determined with this information.
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19
The set of all proteins encoded by the genome is called the:
A)genome.
B)transcriptome.
C)metabolome.
D)proteome.
E)glycome.
A)genome.
B)transcriptome.
C)metabolome.
D)proteome.
E)glycome.
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20
As set of overlapping DNA fragments that form a contiguous stretch of DNA is called a:
A)chromosome.
B)sequence.
C)map.
D)contig.
E)clone.
A)chromosome.
B)sequence.
C)map.
D)contig.
E)clone.
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21
The following microarray data are shown for four human genes.Two samples were hybridized to the microarray in this experiment.One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia.The other sample was cDNA generated from normal, mature neural tissue. - Vertical stripes mean hybridization to cancerous B-cell cDNA only.
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.
The gene sequences contained in spot 2 are present only in the B-cell cancer sample.What is a possible function for this gene, based on the microarray data?
A)This gene may be involved in neurotransmitter release at the synapse since B cells are known to be involved in cell-to-cell signaling.
B)This gene may be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly.
C)This gene is involved in the immune function of the B cell because that is the cell type from which this cancer line was initially derived.
D)This gene may be involved in neurotransmitter release at the synapse since B cells are known to be involved in cell-to-cell signaling, and it may also be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly.
E)This gene may be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly, and it is also involved in the immune function of the B cell because that is the cell type from which this cancer line was initially derived.
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.

A)This gene may be involved in neurotransmitter release at the synapse since B cells are known to be involved in cell-to-cell signaling.
B)This gene may be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly.
C)This gene is involved in the immune function of the B cell because that is the cell type from which this cancer line was initially derived.
D)This gene may be involved in neurotransmitter release at the synapse since B cells are known to be involved in cell-to-cell signaling, and it may also be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly.
E)This gene may be involved in activating cell growth, so it is switched on in the cancer cell line that is growing rapidly, and it is also involved in the immune function of the B cell because that is the cell type from which this cancer line was initially derived.
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22
The following microarray data are shown for four human genes.Two samples were hybridized to the microarray in this experiment.One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia.The other sample was cDNA generated from normal, mature neural tissue. - Vertical stripes mean hybridization to cancerous B-cell cDNA only.
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.
One spot on the microarray contains sequences unique to a gene for an RNA polymerase subunit.One spot contains sequences unique to the skeletal muscle myosin gene.Which spot would be expected to contain the RNA polymerase gene sequences?
A)Spot 1
B)Spot 2
C)Spot 3
D)Spot 4
E)None of the spots
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.

A)Spot 1
B)Spot 2
C)Spot 3
D)Spot 4
E)None of the spots
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23
The largest proportion of human genes are:
A)of unknown function.
B)transcription factors.
C)nucleic acid enzymes.
D)receptors.
E)regulatory molecules.
A)of unknown function.
B)transcription factors.
C)nucleic acid enzymes.
D)receptors.
E)regulatory molecules.
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24
Which of the following statements about eukaryotic genomes is TRUE?
A)A substantial part of the genomes of most multicellular organisms consists of protein-coding sequences.
B)People differ, not only at millions of individual SNPs, but also in the number of copies of many larger segments of the genome.
C)In humans, the number of proteins is approximately equal to the total number of genes.
D)There is a directly proportional relationship between the size of a eukaryotic organism and the number of genes it has.
E)The number of genes among multicellular eukaryotes is related to phenotypic complexity.
A)A substantial part of the genomes of most multicellular organisms consists of protein-coding sequences.
B)People differ, not only at millions of individual SNPs, but also in the number of copies of many larger segments of the genome.
C)In humans, the number of proteins is approximately equal to the total number of genes.
D)There is a directly proportional relationship between the size of a eukaryotic organism and the number of genes it has.
E)The number of genes among multicellular eukaryotes is related to phenotypic complexity.
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25
Which of the following statements about prokaryotic genomes is FALSE?
A)Bacteria can obtain new genetic information through horizontal gene transfer.
B)All prokaryotes have a genome that consists of a single circular chromosome.
C)Prokaryotes with the smallest genomes tend to be in species that occupy restricted habitats, such as bacteria that live inside other organisms.
D)In many sequenced genomes, the function of a significant percentage of genes is currently unknown.
E)The number of genes between species can vary from a few hundred to a few thousand.
A)Bacteria can obtain new genetic information through horizontal gene transfer.
B)All prokaryotes have a genome that consists of a single circular chromosome.
C)Prokaryotes with the smallest genomes tend to be in species that occupy restricted habitats, such as bacteria that live inside other organisms.
D)In many sequenced genomes, the function of a significant percentage of genes is currently unknown.
E)The number of genes between species can vary from a few hundred to a few thousand.
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26
Which of these is FALSE about the ENCODE project?
A)It was a 10-year project.
B)Its purpose was to determine whether noncoding DNAs had a purpose.
C)It concluded that ~80% of the human genome has some function.
D)It concluded that all proteins are equally important.
E)It concluded that some sequences are important for regulation even if they do not encode for genes.
A)It was a 10-year project.
B)Its purpose was to determine whether noncoding DNAs had a purpose.
C)It concluded that ~80% of the human genome has some function.
D)It concluded that all proteins are equally important.
E)It concluded that some sequences are important for regulation even if they do not encode for genes.
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27
This following microarray, comparing normal and cancer cells shows that the: 
A)top part of the panel are miRs overexpressed in cancer.
B)top part of the panel are miRs under-expressed in normal cells.
C)bottom part of the panel are miRs overexpressed in cancer.
D)bottom part of the panel are miRs under expressed in normal cells.
E)top part of the panel are miRs that are both overexpressed in cancer and under-expressed in normal cells.

A)top part of the panel are miRs overexpressed in cancer.
B)top part of the panel are miRs under-expressed in normal cells.
C)bottom part of the panel are miRs overexpressed in cancer.
D)bottom part of the panel are miRs under expressed in normal cells.
E)top part of the panel are miRs that are both overexpressed in cancer and under-expressed in normal cells.
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28
The average gene in the human genome is approximately _____ base pairs in length.
A)700
B)7000
C)17, 000
D)27, 000
E)57, 000
A)700
B)7000
C)17, 000
D)27, 000
E)57, 000
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29
In mass spectrometry, a molecule is ionized and its migration rate in an electrical field is determined.What information does the migration rate tell about the molecule?
A)It indicates mass because small molecules migrate more rapidly than larger molecules.
B)It indicates electrical charge because negatively charged molecules migrate more rapidly than positively charged molecules.
C)It indicates the solubility of the molecule because insoluble molecules form aggregates and migrate much slower than soluble molecules.
D)It indicates the relative abundance of the molecules in the sample.
E)It indicates the complexity of the proteome because greater proteome diversity takes a longer time to process.
A)It indicates mass because small molecules migrate more rapidly than larger molecules.
B)It indicates electrical charge because negatively charged molecules migrate more rapidly than positively charged molecules.
C)It indicates the solubility of the molecule because insoluble molecules form aggregates and migrate much slower than soluble molecules.
D)It indicates the relative abundance of the molecules in the sample.
E)It indicates the complexity of the proteome because greater proteome diversity takes a longer time to process.
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30
A typical prokaryotic genome has:
A)1 million base pairs of DNA, containing 1000 genes.
B)1 million base pairs of DNA, containing a few hundred genes.
C)1 thousand base pairs of DNA, containing a few hundred genes.
D)1 thousand base pairs of DNA, containing a few thousand genes.
E)1 million base pairs of DNA, containing a few million genes.
A)1 million base pairs of DNA, containing 1000 genes.
B)1 million base pairs of DNA, containing a few hundred genes.
C)1 thousand base pairs of DNA, containing a few hundred genes.
D)1 thousand base pairs of DNA, containing a few thousand genes.
E)1 million base pairs of DNA, containing a few million genes.
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31
Which of the following is FALSE, based on this figure? 
A)The majority of human introns are between 1-2 kb.
B)Humans, nematodes, and fruit flies all have introns.
C)Human introns are longer on average than nematode and fly introns.
D)Nematode introns are longer than fruit fly introns.
E)In general, there are few introns bigger than 30 kb.

A)The majority of human introns are between 1-2 kb.
B)Humans, nematodes, and fruit flies all have introns.
C)Human introns are longer on average than nematode and fly introns.
D)Nematode introns are longer than fruit fly introns.
E)In general, there are few introns bigger than 30 kb.
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32
The study of the expression, localization, and interactions of the complete set of proteins found in a given cell is called:
A)genomics.
B)bioinformatics.
C)metagenomics.
D)proteomics.
E)transcriptomics.
A)genomics.
B)bioinformatics.
C)metagenomics.
D)proteomics.
E)transcriptomics.
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33
You are attempting to generate a physical map of the bacteriophage genome using restriction enzyme digestions.You digest the phage DNA with EcoRI or NcoI or both, and observe the fragment sizes listed below.You know that bacteriophage has a linear, not circular, genome.
How many times does the EcoRI enzyme cut the DNA? How many times does the NcoI enzyme cut the DNA?

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34
The following is a microarray that includes genes identified in breast cancer tumors.Why is this gene expression pattern important?
[Reprinted by permission from Macmillan Publishers Ltd.L.J.van'T Veer et al., "Gene expression profiling predicts clinical outcome of breast cancer, " Nature 405:532.© 2002.]
A)It can tell us something about which genes are involved in more invasive or aggressive breast cancer.
B)It can tell us the prognosis of new patients.
C)It can tell us which mutations caused the cancer.
D)It can show us which cDNA is better for diagnosing cancer.
E)It can both tell us something about which genes are involved in more invasive or aggressive breast cancer and tell us the prognosis of new patients.
![<strong>The following is a microarray that includes genes identified in breast cancer tumors.Why is this gene expression pattern important? [Reprinted by permission from Macmillan Publishers Ltd.L.J.van'T Veer et al., Gene expression profiling predicts clinical outcome of breast cancer, Nature 405:532.© 2002.]</strong> A)It can tell us something about which genes are involved in more invasive or aggressive breast cancer. B)It can tell us the prognosis of new patients. C)It can tell us which mutations caused the cancer. D)It can show us which cDNA is better for diagnosing cancer. E)It can both tell us something about which genes are involved in more invasive or aggressive breast cancer and tell us the prognosis of new patients.](https://storage.examlex.com/TB5356/11eaaeec_cbd9_935b_933a_5ba24f84f8d1_TB5356_00.jpg)
A)It can tell us something about which genes are involved in more invasive or aggressive breast cancer.
B)It can tell us the prognosis of new patients.
C)It can tell us which mutations caused the cancer.
D)It can show us which cDNA is better for diagnosing cancer.
E)It can both tell us something about which genes are involved in more invasive or aggressive breast cancer and tell us the prognosis of new patients.
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35
How are proteins analyzed in mass spectrometry?
A)The mass spectrometer shows the sequence of the individual amino acids.
B)The mass spectrometer compares the profile of peaks to its database of other proteins, looking for a match.
C)The mass spectrometer fragments the protein to small pieces.
D)The mass spectrometer shows the m/z ratio of each protein, and each protein has a different m/z ratio.
E)The mass spectrometer can predict DNA sequences of proteins.
A)The mass spectrometer shows the sequence of the individual amino acids.
B)The mass spectrometer compares the profile of peaks to its database of other proteins, looking for a match.
C)The mass spectrometer fragments the protein to small pieces.
D)The mass spectrometer shows the m/z ratio of each protein, and each protein has a different m/z ratio.
E)The mass spectrometer can predict DNA sequences of proteins.
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36
How much of the human genome codes for proteins?
A)Less than 2%
B)Approximately 24%
C)Approximately 50%
D)Approximately 74%
E)More than 82%
A)Less than 2%
B)Approximately 24%
C)Approximately 50%
D)Approximately 74%
E)More than 82%
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37
The following microarray data are shown for four human genes.Two samples were hybridized to the microarray in this experiment.One sample was cDNA generated from mRNA of a tissue culture cell line that was originally derived from a B-cell leukemia.The other sample was cDNA generated from normal, mature neural tissue. - Vertical stripes mean hybridization to cancerous B-cell cDNA only.
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.
One spot on the microarray contains sequences unique to a gene for an RNA polymerase subunit.One spot contains sequences unique to the skeletal muscle myosin gene.Which spot would be expected to contain the myosin gene sequences?
A)Spot 1
B)Spot 2
C)Spot 3
D)Spot 4
E)None of the spots
- Horizontal stripes mean hybridization to normal neural cell cDNA only.
- Completely filled in means hybridization to both cDNA types.
- Blank means hybridization to neither cell's cDNA.

A)Spot 1
B)Spot 2
C)Spot 3
D)Spot 4
E)None of the spots
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38
Explain why genetic and physical map distances may differ in relative distances between two genes on a chromosome.
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39
After genomics, scientists turned to proteomics because:
A)even though DNA sequences are important, they do not tell the whole story of the organism.
B)they figured out everything there was to know about DNA.
C)proteins were deemed more important than DNA.
D)proteins are easier to study.
E)there are less proteins than there are genes, and they are thus easier to study.
A)even though DNA sequences are important, they do not tell the whole story of the organism.
B)they figured out everything there was to know about DNA.
C)proteins were deemed more important than DNA.
D)proteins are easier to study.
E)there are less proteins than there are genes, and they are thus easier to study.
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40
In the figure, it is MOST likely that: 
A)the top 18 miRs should be turned off to treat the cancer.
B)the bottom 15 miRs should be turned on to treat the cancer.
C)the top 18 miRs should be turned on to treat the cancer.
D)the top 18 miRs should be turned off and the bottom 15 miRs should be turned on to treat the cancer.
E)both the bottom 15 miRs and the top 18 miRs should be turned on to treat the cancer.

A)the top 18 miRs should be turned off to treat the cancer.
B)the bottom 15 miRs should be turned on to treat the cancer.
C)the top 18 miRs should be turned on to treat the cancer.
D)the top 18 miRs should be turned off and the bottom 15 miRs should be turned on to treat the cancer.
E)both the bottom 15 miRs and the top 18 miRs should be turned on to treat the cancer.
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41
What is synthetic biology, and why is it a concern?
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42
How has the study of microbiomics shed light on the understanding of obesity?
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43
How are SNPs used to search for genes causing disease?
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44
If a DNA sequence of a gene is known, one can synthesize the protein and find out its function.What are the disadvantages of this approach, and what other approaches can be taken?
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45
Describe one major difference in the organization or content of prokaryotic and eukaryotic genomes.
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46
The transcription factor Twist helps specify development of the mesoderm in the Drosophila embryo.You wish to identify other genes that Twist (as a transcription factor)switches on or off.You have wild-type embryos available, as well as embryos that are mutant for the Twist gene.Design an experiment using microarrays that could allow you to identify genes switched on or off by the Twist transcription factor.
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47
Compare the fields of structural, functional, and comparative genomics.What is the purpose of each?
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48
How has metagenomics of ocean samples contributed to our understanding of energy in biological systems?
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49
The transcriptome of a genome contains more components than the proteome.Explain why this is TRUE.
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50
What is the main difference between map-based sequencing and whole genome shotgun sequencing?
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