Deck 15: Genes and How They Work
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Question
Unlock Deck
Sign up to unlock the cards in this deck!
Unlock Deck
Unlock Deck
1/67
Play
Full screen (f)
Deck 15: Genes and How They Work
1
A codon is composed of how many bases?
A)one
B)two
C)three
D)four
A)one
B)two
C)three
D)four
C
2
The polypeptide-making organelles, which consist of protein combined with RNA, are called
A)ribosomes.
B)Golgi bodies.
C)lysosomes.
D)centrosomes.
A)ribosomes.
B)Golgi bodies.
C)lysosomes.
D)centrosomes.
A
3
Messenger RNA molecules contain information that is used to synthesize ___________.
A)nucleotides
B)amino acids
C)polypeptides
D)fatty acids
A)nucleotides
B)amino acids
C)polypeptides
D)fatty acids
C
4
Most eukaryotic genes contain noncoding sequences called ________ that are interspersed with the coding sequences.
A)introns
B)exons
C)codons
D)spacers
A)introns
B)exons
C)codons
D)spacers
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
5
During ________, a ribosome assembles a polypeptide whose amino acid sequence is specified by the nucleotide sequence in a molecule of mRNA.
A)transcription
B)translation
C)replication
D)posttranscriptional modification
A)transcription
B)translation
C)replication
D)posttranscriptional modification
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
6
Which base in an anticodon will pair with the base adenine in a codon?
A)thymine
B)cytosine
C)guanine
D)uracil
A)thymine
B)cytosine
C)guanine
D)uracil
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
7
To begin transcription, RNA polymerase must bind to a segment of DNA called the ____.
A)initiation site
B)primer
C)inducer
D)promoter
A)initiation site
B)primer
C)inducer
D)promoter
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
8
DNA affects the traits of an organism by providing the instructions for synthesizing _______.
A)proteins
B)nucleotides
C)codons
D)amino acids
A)proteins
B)nucleotides
C)codons
D)amino acids
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
9
Gene ________ refers to the combined processes of transcription and translation.
A)expression
B)replication
C)modification
D)regulation
A)expression
B)replication
C)modification
D)regulation
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
10
During transcription of mRNA in eukaryotes, some sequences are cut out of the primary transcript and the remaining sequences are joined together.This processing of mRNA is called _________.
A)termination
B)translation
C)splicing
D)capping
A)termination
B)translation
C)splicing
D)capping
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
11
To remove noncoding sequences in the pre-mRNA of eukaryotes, multiple snRNPs combine with proteins to form a larger complex called the ___________ .
A)5' cap
B)introsome
C)ribosome
D)spliceosome
A)5' cap
B)introsome
C)ribosome
D)spliceosome
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
12
The connection that exists between genes and hereditary traits is based on using the information encoded in genes to synthesize
A)codons.
B)nucleotides.
C)proteins.
D)histones.
A)codons.
B)nucleotides.
C)proteins.
D)histones.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
13
The strand of DNA that is not transcribed is called the ______ strand.
A)coding
B)non-coding
C)template
D)complementary
A)coding
B)non-coding
C)template
D)complementary
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
14
During translation, amino acids are carried to the ribosome by
A)mRNA.
B)tRNA.
C)snRNA.
D)rRNA.
A)mRNA.
B)tRNA.
C)snRNA.
D)rRNA.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
15
Gene expression involves two phases, ___________ and translation.
A)replication
B)transcription
C)initiation
D)condensation
A)replication
B)transcription
C)initiation
D)condensation
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
16
Crick and his colleagues proposed that the genetic code consists of a series of blocks of information, called ______, each corresponding to one amino acid in an encoded protein.
A)alleles
B)codons
C)genes
D)polypeptides
A)alleles
B)codons
C)genes
D)polypeptides
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
17
The sequence of nucleotides in a DNA molecule is called the ________ code.
A)protein
B)ribosomal
C)translation
D)genetic
A)protein
B)ribosomal
C)translation
D)genetic
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
18
In eukaryotes, translation takes place
A)on the plasma membrane.
B)inside the nucleus.
C)on ribosomes.
D)on the nuclear membrane.
A)on the plasma membrane.
B)inside the nucleus.
C)on ribosomes.
D)on the nuclear membrane.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
19
Both DNA and RNA are made up of building blocks known as
A)nucleotides.
B)nucleic acids.
C)amino acids.
D)genes.
A)nucleotides.
B)nucleic acids.
C)amino acids.
D)genes.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
20
During _______, RNA polymerase synthesizes a molecule of RNA using DNA as a template.
A)mRNA splicing
B)translation
C)transcription
D)gene sequencing
A)mRNA splicing
B)translation
C)transcription
D)gene sequencing
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
21
In eukaryotic cells, transcription occurs
A)on the surface of the nuclear membrane.
B)on ribosomes.
C)on spliceosomes.
D)inside the nucleus.
A)on the surface of the nuclear membrane.
B)on ribosomes.
C)on spliceosomes.
D)inside the nucleus.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
22
Which statement about the genetic code is false?
A)There is no punctuation or spacing between codons.
B)Nucleotides are always read in groups of three.
C)Every codon codes for one amino acid.
D)Some amino acids are specified by more than one codon.
A)There is no punctuation or spacing between codons.
B)Nucleotides are always read in groups of three.
C)Every codon codes for one amino acid.
D)Some amino acids are specified by more than one codon.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
23
If the sequence of bases in the template strand of a DNA molecule is 3' ATCGCTCC 5', what is the sequence of bases in the RNA that is transcribed from this molecule?
A)3' UAGCGAGG 5'
B)3' TAGCGAGG 5'
C)5' UAGCGAGG 3'
D)5' TAGCGAGG 3'
A)3' UAGCGAGG 5'
B)3' TAGCGAGG 5'
C)5' UAGCGAGG 3'
D)5' TAGCGAGG 3'
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
24
The A, P, and E sites are progressively occupied by amino acids being assembled into a polypeptide.These sites are part of
A)DNA.
B)the large ribosomal subunit.
C)mRNA.
D)tRNA.
A)DNA.
B)the large ribosomal subunit.
C)mRNA.
D)tRNA.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
25
The Central Dogma of biology can be stated as
A)proteins RNA DNA.
B)RNA DNA proteins.
C)DNA proteins RNA.
D)DNA RNA proteins.
A)proteins RNA DNA.
B)RNA DNA proteins.
C)DNA proteins RNA.
D)DNA RNA proteins.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
26
During translation, uncharged tRNA molecules leave the ribosome from the _________ site.
A)E
B)P
C)A
D)termination
A)E
B)P
C)A
D)termination
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
27
The genetic code uses _________ nucleotide(s) to specify one amino acid.
A)one
B)two
C)three
D)four
A)one
B)two
C)three
D)four
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
28
Codons that serve as "stop" signals for translation are recognized by
A)tRNA.
B)release factors.
C)anticodons.
D)translation terminators.
A)tRNA.
B)release factors.
C)anticodons.
D)translation terminators.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
29
What is the first step during transcription initiation in prokaryotes?
A)the transcription bubble is formed
B)RNA polymerase binds to the promoter
C)the DNA double helix is unwound
D)RNA polymerase synthesizes a short primer
A)the transcription bubble is formed
B)RNA polymerase binds to the promoter
C)the DNA double helix is unwound
D)RNA polymerase synthesizes a short primer
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
30
During translation, the nucleotides that make up the mRNA are read in groups of three.These groups are called
A)codons.
B)anticodons.
C)exons.
D)introns.
A)codons.
B)anticodons.
C)exons.
D)introns.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
31
Ribosomes are complex aggregates of
A)RNA and DNA.
B)RNA and proteins.
C)RNA and sugars.
D)DNA and proteins.
A)RNA and DNA.
B)RNA and proteins.
C)RNA and sugars.
D)DNA and proteins.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
32
The location of translation in prokaryotic cells is
A)in the nucleoid.
B)on ribosomes.
C)on the plasma membrane.
D)on mesosomes.
A)in the nucleoid.
B)on ribosomes.
C)on the plasma membrane.
D)on mesosomes.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
33
Eukaryotic mRNA molecules may contain non-coding sequences that must be removed before translation.These are called
A)anticodons.
B)introns.
C)exons.
D)nucleosomes.
A)anticodons.
B)introns.
C)exons.
D)nucleosomes.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
34
During translation in prokaryotes, formation of the initiation complex requires all of the following except
A)a small ribosomal subunit.
B)mRNA.
C)tRNA charged with N-formylmethionine.
D)RNA polymerase.
A)a small ribosomal subunit.
B)mRNA.
C)tRNA charged with N-formylmethionine.
D)RNA polymerase.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
35
The template strand of a DNA segment that codes for mRNA has the sequence: ATGCGT.Which tRNA anticodons would pair with the mRNA that is coded for by this sequence?
A)AUG CGU.
B)ATG CGT.
C)UAC GCA.
D)UAG CGU.
A)AUG CGU.
B)ATG CGT.
C)UAC GCA.
D)UAG CGU.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
36
Although 61 different codons code for amino acids, cells contain fewer than 61 different tRNAs.Why?
A)Because the 5' base on the tRNA anticodon has some flexibility (wobble); thus, some tRNA anticodons can pair with more than one mRNA codon.
B)Although 61 different codons code for amino acids, any given cell contains fewer than 61.
C)Because the 5' base on the mRNA codon has some flexibility (wobble); thus, some mRNA codons can pair with more than one tRNA anticodon.
D)Because each amino acid is coded for by just one codon.
A)Because the 5' base on the tRNA anticodon has some flexibility (wobble); thus, some tRNA anticodons can pair with more than one mRNA codon.
B)Although 61 different codons code for amino acids, any given cell contains fewer than 61.
C)Because the 5' base on the mRNA codon has some flexibility (wobble); thus, some mRNA codons can pair with more than one tRNA anticodon.
D)Because each amino acid is coded for by just one codon.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
37
In eukaryotes, pre-mRNA processing may involve all of the following except
A)removal of exons from the pre-mRNA.
B)addition of a 5' cap.
C)addition of a 3' poly-A tail
D)pre-mRNA splicing by the spliceosome.
A)removal of exons from the pre-mRNA.
B)addition of a 5' cap.
C)addition of a 3' poly-A tail
D)pre-mRNA splicing by the spliceosome.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
38
In prokaryotes, the form of RNA polymerase that can accurately initiate synthesis of RNA is called
A)the holoenzyme.
B)the core polymerase.
C)RNA polymerase II.
D)RNA polymerase III.
A)the holoenzyme.
B)the core polymerase.
C)RNA polymerase II.
D)RNA polymerase III.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
39
Specific amino acids are attached to tRNA molecules by
A)aminoacyl-tRNA synthetases.
B)hydrogen bonds.
C)anticodons.
D)deactivating enzymes.
A)aminoacyl-tRNA synthetases.
B)hydrogen bonds.
C)anticodons.
D)deactivating enzymes.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
40
Ribosome movement along the mRNA is called
A)transcription.
B)initiation.
C)replication.
D)translocation.
A)transcription.
B)initiation.
C)replication.
D)translocation.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
41
You are working to characterize a novel protein in mice.Analysis shows that high levels of the primary transcript that codes for this protein are found in tissue from the brain, muscle, liver, and pancreas.However, an antibody that recognizes the C-terminal portion of the protein indicates that the protein is present in brain, muscle, and liver, but not in the pancreas.What is the most likely explanation for this result?
A)The gene that codes for this protein is not transcribed in the pancreas.
B)The transcript is immediately degraded in the pancreas.
C)The transcript receives a 5' cap in the pancreas.
D)Alternative splicing in the pancreas creates a protein missing the domain that the antibody recognizes.
A)The gene that codes for this protein is not transcribed in the pancreas.
B)The transcript is immediately degraded in the pancreas.
C)The transcript receives a 5' cap in the pancreas.
D)Alternative splicing in the pancreas creates a protein missing the domain that the antibody recognizes.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
42
Eukaryotes have ____ type(s) of RNA polymerase.
A)two
B)three
C)four
D)one
A)two
B)three
C)four
D)one
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
43
What is the likely consequence of a mutation that alters the branch point within an intron?
A)no effect, since introns are not expressed
B)failure to form a lariat
C)failure of snRNPs to recognize the 5' end of intron
D)no exon shuffling
A)no effect, since introns are not expressed
B)failure to form a lariat
C)failure of snRNPs to recognize the 5' end of intron
D)no exon shuffling
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
44
What would happen if snRNPs did not recognize the branch point within an intron?
A)A lariat would not form.
B)snRNPs would not base-pair with the 5' end of the intron.
C)A 3' poly A tail would not be added to the transcript.
D)A 5' cap would not be added to the transcript.
A)A lariat would not form.
B)snRNPs would not base-pair with the 5' end of the intron.
C)A 3' poly A tail would not be added to the transcript.
D)A 5' cap would not be added to the transcript.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
45
In eukaryotes, each type of RNA polymerase recognizes a different
A)start codon
B)stop codon
C)promoter
D)release factor
A)start codon
B)stop codon
C)promoter
D)release factor
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
46
Which molecule combines with proteins to form both the large and small ribosomal subunits?
A)DNA
B)RNA polymerase
C)miRNA
D)rRNA
A)DNA
B)RNA polymerase
C)miRNA
D)rRNA
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
47
Initiation of transcription differs from initiation of DNA replication in several ways.One difference is that initiation of transcription does not require
A)a promoter.
B)enzymes.
C)a primer.
D)a DNA template strand.
A)a promoter.
B)enzymes.
C)a primer.
D)a DNA template strand.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
48
In prokaryotes, the RNA polymerase holoenzyme consists of
A)the core polymerase plus two alpha subunits.
B)the core polymerase plus two beta subunits.
C)the core polymerase plus two alpha subunits, two beta subunits, and a sigma subunit.
D)the core polymerase plus a sigma subunit.
A)the core polymerase plus two alpha subunits.
B)the core polymerase plus two beta subunits.
C)the core polymerase plus two alpha subunits, two beta subunits, and a sigma subunit.
D)the core polymerase plus a sigma subunit.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
49
Cells conserve energy and resources by making active proteins only when they are needed.If a protein is not needed, which of the following methods of control would be the most energy-efficient?
A)block transcription
B)degrade the mRNA after it is made
C)prevent translation of the mRNA
D)degrade the protein after it is made
A)block transcription
B)degrade the mRNA after it is made
C)prevent translation of the mRNA
D)degrade the protein after it is made
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
50
During protein synthesis in eukaryotes, what happens during RNA splicing?
A)The product of translation, called the primary transcript, is cut and some pieces are joined back together to form the mature mRNA.
B)The product of transcription, called the primary transcript, is cut and some pieces are joined back together to form the mature tRNA.
C)The product of transcription, called the secondary transcript, is cut and some pieces are joined back together to form the mature mRNA.
D)The product of transcription, called the primary transcript, is cut and some pieces are joined back together to form the mature mRNA.
A)The product of translation, called the primary transcript, is cut and some pieces are joined back together to form the mature mRNA.
B)The product of transcription, called the primary transcript, is cut and some pieces are joined back together to form the mature tRNA.
C)The product of transcription, called the secondary transcript, is cut and some pieces are joined back together to form the mature mRNA.
D)The product of transcription, called the primary transcript, is cut and some pieces are joined back together to form the mature mRNA.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
51
During protein synthesis in eukaryotes, which molecule passes from the nucleus to the cytoplasm and specifies the sequence of amino acids in the new polypeptide?
A)DNA
B)RNA polymerase
C)mRNA
D)rRNA
A)DNA
B)RNA polymerase
C)mRNA
D)rRNA
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
52
In eukaryotes, the 3' poly-A tail is attached to
A)poly-A polymerase.
B)mRNA.
C)tRNA.
D)the ribosome.
A)poly-A polymerase.
B)mRNA.
C)tRNA.
D)the ribosome.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
53
During the splicing reaction, the intron-exon junctions are recognized by
A)snRNPs.
B)miRNAs.
C)SRP RNAs.
D)the lariat.
A)snRNPs.
B)miRNAs.
C)SRP RNAs.
D)the lariat.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
54
Given the sentence "THE FAT CAT ATE THE RED RAT," which of the following would represent a frameshift mutation?
A)THE FAT RAT ATE THE RED CAT
B)THE CAT ATE THE RED RAT
C)THE FAC ATA TET HER EDR AT
D)THE FAT CAT ATE THE RED RAT
A)THE FAT RAT ATE THE RED CAT
B)THE CAT ATE THE RED RAT
C)THE FAC ATA TET HER EDR AT
D)THE FAT CAT ATE THE RED RAT
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
55
Eukaryotic pre-mRNA molecules are modified
A)in the cytoplasm.
B)at the ribosome.
C)inside the nucleus.
D)as they pass through the nuclear membrane.
A)in the cytoplasm.
B)at the ribosome.
C)inside the nucleus.
D)as they pass through the nuclear membrane.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
56
What is required for formation of the transcription initiation complex in eukaryotes?
A)binding of a transcription factor to the TATA box, followed by recruitment of additional transcription factors and recruitment of RNA polymerase II
B)binding of a transcription factor to the transcription bubble, followed by recruitment of additional transcription factors and recruitment of RNA polymerase III
C)binding of the sigma subunit to the start site followed by recruitment of RNA polymerase II
D)binding of RNA polymerase II to the TATA box, followed by recruitment of transcription factors
A)binding of a transcription factor to the TATA box, followed by recruitment of additional transcription factors and recruitment of RNA polymerase II
B)binding of a transcription factor to the transcription bubble, followed by recruitment of additional transcription factors and recruitment of RNA polymerase III
C)binding of the sigma subunit to the start site followed by recruitment of RNA polymerase II
D)binding of RNA polymerase II to the TATA box, followed by recruitment of transcription factors
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
57
The mutation responsible for Huntington's disease is a
A)missense mutation.
B)nonsense mutation.
C)frameshift mutation.
D)triplet repeat expansion mutation.
A)missense mutation.
B)nonsense mutation.
C)frameshift mutation.
D)triplet repeat expansion mutation.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
58
Within the transcription bubble, the 9 most recently added nucleotides in the newly synthesized RNA strand temporarily form a helix with the template DNA strand.How might transcription be affected if helix formation did not occur?
A)Rewinding the DNA molecule would be inhibited.
B)Unwinding the DNA molecule would be inhibited.
C)The position of the 5' end of the RNA would be unstable, inhibiting elongation.
D)The position of the 3' end of the RNA would be unstable, inhibiting elongation.
A)Rewinding the DNA molecule would be inhibited.
B)Unwinding the DNA molecule would be inhibited.
C)The position of the 5' end of the RNA would be unstable, inhibiting elongation.
D)The position of the 3' end of the RNA would be unstable, inhibiting elongation.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
59
Why are there fewer tRNA anticodons than the 61 needed to match each mRNA codon that codes for an amino acid?
A)There is some flexibility in pairing between the 5' base of the codon and the 3' base of the anticodon.
B)There is some flexibility in pairing between the middle base of the codon and the middle base of the anticodon.
C)There is some flexibility in pairing between the 3' base of the codon and the 5' base of the anticodon.
D)There is some flexibility in pairing between all 3 bases of the codon and all 3 bases of the anticodon.
A)There is some flexibility in pairing between the 5' base of the codon and the 3' base of the anticodon.
B)There is some flexibility in pairing between the middle base of the codon and the middle base of the anticodon.
C)There is some flexibility in pairing between the 3' base of the codon and the 5' base of the anticodon.
D)There is some flexibility in pairing between all 3 bases of the codon and all 3 bases of the anticodon.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
60
Two 6-base sequences are present in bacterial promoters: TATAAT (located 10 nt upstream from the start site) and TTGACA (located 35 nt upstream from the start site).What is the significance of the fact that these two base sequences are different?
A)Binding sites for both the holoenzyme and ATP are provided.
B)Both the location of the start site and the direction of transcription can be established.
C)Binding sites for both the core polymerase and holoenzyme are provided.
D)The transcription bubble can be properly formed.
A)Binding sites for both the holoenzyme and ATP are provided.
B)Both the location of the start site and the direction of transcription can be established.
C)Binding sites for both the core polymerase and holoenzyme are provided.
D)The transcription bubble can be properly formed.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
61
A bacterial cell has a nonsense mutation that prevents it from producing a functional sigma subunit for RNA polymerase.Inability to synthesize a functional sigma subunit would have the most direct effect on
A)transcription initiation.
B)transcription elongation.
C)transcription termination.
D)translation initiation.
A)transcription initiation.
B)transcription elongation.
C)transcription termination.
D)translation initiation.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
62
A gene that codes for a protein was removed from a eukaryotic cell and inserted into a prokaryotic cell.Although the gene was successfully transcribed and translated, it produced a different protein than it produced in the eukaryotic cell.What is the most likely explanation?
A)There are slight differences in the genetic code for prokaryotes and eukaryotes.
B)Unlike eukaryotes, which have three different RNA polymerases, prokaryotes have a single RNA polymerase.
C)Eukaryotic genes often contain introns while prokaryotic genes do not.
D)Eukaryotic transcripts have a 5' cap while prokaryotic transcripts do not.
A)There are slight differences in the genetic code for prokaryotes and eukaryotes.
B)Unlike eukaryotes, which have three different RNA polymerases, prokaryotes have a single RNA polymerase.
C)Eukaryotic genes often contain introns while prokaryotic genes do not.
D)Eukaryotic transcripts have a 5' cap while prokaryotic transcripts do not.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
63
You are attempting to synthesize rRNA in a test tube using DNA isolated from mouse cells.In addition to the template DNA, ribonucleotides, and the necessary transcription factors, you should also add _________ to the test tube.
A)poly-A polymerase
B)RNA polymerase III
C)RNA polymerase II
D)RNA polymerase I
A)poly-A polymerase
B)RNA polymerase III
C)RNA polymerase II
D)RNA polymerase I
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
64
How does DNA polymerase differ from RNA polymerase?
A)Only RNA polymerase adds new nucleotides to the 3' end of a growing chain.
B)Only RNA polymerase requires a primer.
C)Only DNA polymerase uses a template DNA strand to direct synthesis of a new nucleotide strand.
D)Only DNA polymerase has a proofreading ability.
A)Only RNA polymerase adds new nucleotides to the 3' end of a growing chain.
B)Only RNA polymerase requires a primer.
C)Only DNA polymerase uses a template DNA strand to direct synthesis of a new nucleotide strand.
D)Only DNA polymerase has a proofreading ability.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
65
How would a large chromosomal inversion affect the expression of a gene if the gene is located between the two break points but no breaks occur within the gene?
A)The inversion would probably have no effect on gene expression.
B)The gene would not be transcribed because it would be oriented in the wrong direction.
C)The gene would be transcribed in the 3' to 5' direction.
D)The gene would be transcribed normally but the mRNA would be translated in the 3' to 5' direction.
A)The inversion would probably have no effect on gene expression.
B)The gene would not be transcribed because it would be oriented in the wrong direction.
C)The gene would be transcribed in the 3' to 5' direction.
D)The gene would be transcribed normally but the mRNA would be translated in the 3' to 5' direction.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
66
A scientist makes three artificial mRNA strands: (x) 5' AAAUUUAAAUUUAAAUUUAAAUUUAAA 3'
(y) 5' UUUCCCUUUCCCUUUCCCUUUCCCUUU 3'
(z) 5' AUAUAUAUAUAUAUAUAUAUAUAUAU 3'
When he analyzes the polypeptides produced, he finds that:
X produces a polypeptide that is 50% phenylalanine and 50%lysine.y produces a polypeptide that is 50% phenylalanine and 50% proline.z produces a polypeptide that is 50% isoleucine and 50% tyrosine.Based on these results only, the best conclusion to make is that
A)AUA codes for isoleucine
B)AAA codes for phenylalanine
C)AAA codes for lysine
D)AAA codes for lysine and AUA codes for isoleucine
(y) 5' UUUCCCUUUCCCUUUCCCUUUCCCUUU 3'
(z) 5' AUAUAUAUAUAUAUAUAUAUAUAUAU 3'
When he analyzes the polypeptides produced, he finds that:
X produces a polypeptide that is 50% phenylalanine and 50%lysine.y produces a polypeptide that is 50% phenylalanine and 50% proline.z produces a polypeptide that is 50% isoleucine and 50% tyrosine.Based on these results only, the best conclusion to make is that
A)AUA codes for isoleucine
B)AAA codes for phenylalanine
C)AAA codes for lysine
D)AAA codes for lysine and AUA codes for isoleucine
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck
67
You are studying an individual with very low levels of insulin in her blood.Further analysis indicates that cells of her pancreas are producing normal levels of this protein, but it is accumulating in the cytoplasm rather than being secreted from the cells.Which hypothesis makes the most sense to explain this observation?
A)A small deletion has removed the nucleotides that code for the signal sequence at the amino terminus of the protein.
B)A missense mutation has caused premature termination during translation of this protein.
C)A chromosomal segment that includes the gene for insulin has been inverted.
D)A two-base deletion near the middle of the gene has altered the reading frame during translation of the protein.
A)A small deletion has removed the nucleotides that code for the signal sequence at the amino terminus of the protein.
B)A missense mutation has caused premature termination during translation of this protein.
C)A chromosomal segment that includes the gene for insulin has been inverted.
D)A two-base deletion near the middle of the gene has altered the reading frame during translation of the protein.
Unlock Deck
Unlock for access to all 67 flashcards in this deck.
Unlock Deck
k this deck