A) gene alignment
B) intron shuffling
C) exon shuffling
D) peptide shuffling
E) exon splicing
G) A) and D)
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A) look for similar gene sequences of known function
B) examine the putative protein structure
C) perform a gene knockout
D) find a pseudogene for the gene
E) determine the amino acid sequence
G) A) and E)
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A) Alternative splicing allows for different transcripts to be created by the same gene sequence.
B) The researcher's sample was contaminated.
C) Exon shuffling allows for different transcripts to be created by the same gene sequence.
D) Intron shuffling allows for different transcripts to be created by the same gene sequence.
E) It is not possible; there was an error in the alignment process.
G) A) and D)
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A) Red and green are two of the primary colors, and if both cDNAs hybridize to the spot, the combination of the two will produce yellow light.
B) The yellow light comes from the laser, and a yellow spot indicates that neither cDNA hybridized to that spot on the microarray.
C) The over expression of one cDNA relative to the other will skew the color pattern of the spot on the microarray, resulting in the yellow color.
D) The color choice was an arbitrary decision by the artist and doesn't reflect how the process actually works.
E) Yellow spots are those that contain no DNA probes on the chip.
G) B) and D)
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A) is a pseudogene
B) encodes the same hemoglobin subunit
C) is inactive because it is in a different species
D) is a regulatory sequence
E) encodes a respiratory enzyme
G) C) and D)
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A) refers to the analysis of the entire protein content of a cell
B) refers to the analysis of all the DNA of a species
C) looks only at plasmids
D) studies mRNA levels
E) uses DNA chips
G) A) and E)
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A) Archaea only
B) Bacteria only
C) Eukarya only
D) Archaea and Eukarya
E) Bacteria and Archaea
G) A) and D)
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A) transposition
B) alternative splicing
C) intron shuffling
D) intron splicing
E) open reading frame shuffling
G) All of the above
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A) DNA is broken into many random, overlapping fragments that are sequenced and then assembled using computer algorithms
B) whole chromosomes are sequencedintact from the 3'end to the 5' end
C) whole chromosomes are sequenced intactfrom the 5'end to the 3' end
D) DNA is broken into a few, non-overlapping fragments that can be read directly by computer algorithms
E) DNA is broken into individual nucleotides that are sequenced and then assembled using computer algorithms
G) C) and D)
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A) We have not yet identified all of the open reading frames in the human genome.
B) We have not yet fully sequenced the human genome.
C) mRNA processing mechanisms allowmultiple proteins to be produced from a single DNA sequence.
D) Some noncoding DNA sequences encode proteins.
E) There has been a gross over-estimation of the number of proteins produced in humans.
G) A) and D)
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A) Males have 46 chromosomes; females have 45.
B) Males have 22 pairs of autosomal chromosomes; females have 21 pairs of autosomal chromosomes.
C) Males have 23 pairs of chromosomes plus 1 Y chromosome; females have 23 pairs of chromosomes.
D) Males have 24 different chromosomes; females have 23 different chromosomes
E) Males have an X chromosome; females do not.
G) A) and B)
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A) The use of embryonic stem cells for gene therapy.
B) The discovery of a gene that predicts an unfavorable drug response in an individual, for whom there are no alternative treatments available.
C) The use of vaccination to prevent diseases.
D) The use of blood transfusions to treat diseases.
E) The introduction of foreign genes to restore loss of vital gene function in a patient with an infectious disease.
G) A) and D)
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A) Microarrays let us study different cellular structures under different conditions.
B) Microarrays let us directly measure protein expression in individual cells.
C) Microarrayslet us identify which portions of a genome were being expressed in a cell at a particular time.
D) Microarrays let us identify which DNA sequences are present in a particular cell type under certain conditions.
E) Microarrays let us identify which portions of a genome serve as regulatory sequences.
G) B) and C)
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A) 2%
B) 20%
C) 30%
D) 50%
E) 85%
G) B) and C)
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A) the 5 ¢ untranslated region and the ORF only
B) the ORF only
C) the ORF and 3 ¢ untranslated region only
D) the 5 ¢ and 3 ¢ untranslated regions only
E) the ORF, 5 ¢ untranslated region, and 3 ¢ untranslated region
G) C) and D)
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A) drug metabolism genes
B) body size
C) blood volume
D) trial and error methods
E) drug interactions
G) C) and D)
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A) a protein coding sequence plus associated regulatory sequences
B) the sequence between and including a start codon and an stop codon
C) the sequence between and including a start codon and an stop codon, minus the introns
D) the sequence between and including a start codon and an stop codon, minus the exons
E) the sequence between a start codon and stop codon in prokaryotes only
G) B) and E)
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