Bsc 1010L Ch 15

Front 1

1. DNA affects the traits of an organism by providing the instructions for synthesizing _______.
A. proteins
B. nucleotides
C. codons
D. amino acids

Back 1

A. proteins

Front 2

2. Gene expression involves two phases, ___________ and translation.
A. replication
B. transcription
C. initiation
D. condensation

Back 2

B. transcription

Front 3

3. Messenger RNA molecules contain information that is used to synthesize ___________.
A. nucleotides
B. amino acids
C. polypeptides
D. fatty acids

Back 3

C. polypeptides

Front 4

4. To begin transcription, RNA polymerase must bind to a segment of DNA called the ____.
A. initiation site
B. primer
C. inducer
D. promoter
E. transcription bubble

Back 4

D. promoter

Front 5

5. The strand of DNA that is not transcribed is called the ______ strand.
A. coding
B. non-coding
C. template
D. complementary

Back 5

A. coding

Front 6

6. 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

Back 6

B. translation

Front 7

7. The sequence of nucleotides in a DNA molecule is called the ________ code.
A. protein
B. ribosomal
C. translation
D. genetic
E. amino acid

Back 7

D. genetic

Front 8

8. 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

Back 8

B. codons

Front 9

9. Gene ________ refers to the combined processes of transcription and translation.
A. expression
B. replication
C. modification
D. regulation

Back 9

A. expression

Front 10

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
E. elongation

Back 10

C. splicing

Front 11

11. Most eukaryotic genes contain noncoding sequences called ________ that are interspersed with the coding sequences.
A. introns
B. exons
C. codons
D. spacers
E. spliceosomes

Back 11

A. introns

Front 12

12. 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
E. 3' poly-A tail

Back 12

D. spliceosome

Front 13

13. 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.
E. complementary bases.

Back 13

C. proteins.

Front 14

14. Both DNA and RNA are made up of building blocks known as
A. nucleotides.
B. nucleic acids.
C. amino acids.
D. genes.
E. codons.

Back 14

A. nucleotides.

Front 15

15. The "one-gene/one-enzyme" hypothesis was proposed by
A. Watson and Crick.
B. Griffith.
C. Garrod.
D. Franklin.
E. Beadle and Tatum.

Back 15

E. Beadle and Tatum.

Front 16

16. The polypeptide-making organelles, which consist of protein combined with RNA, are called
A. ribosomes.
B. Golgi bodies.
C. lysosomes.
D. centrosomes.
E. mitochondria.

Back 16

A. ribosomes.

Front 17

17. During translation, amino acids are carried to the ribosome by
A. mRNA.
B. tRNA.
C. snRNA.
D. rRNA.
E. miRNA.

Back 17

B. tRNA.

Front 18

18. During _______, RNA polymerase synthesizes a molecule of RNA using DNA as a template.
A. mRNA splicing
B. translation
C. transcription
D. gene sequencing
E. termination

Back 18

C. transcription

Front 19

19. Which base in an anticodon will pair with the base adenine in a codon?
A. thymine
B. cytosine
C. guanine
D. uracil

Back 19

D. uracil

Front 20

20. A codon is composed of how many bases?
A. one
B. two
C. three
D. four
E. 64

Back 20

C. three

Front 21

21. In eukaryotes, translation takes place
A. on the plasma membrane.
B. inside the nucleus.
C. on ribosomes.
D. on the nuclear membrane.
E. on spliceosomes.

Back 21

C. on ribosomes.

Front 22

22. Ribosomes are complex aggregates of
A. RNA and DNA.
B. RNA and proteins.
C. RNA and sugars.
D. DNA and proteins.
E. nucleosomes and RNA.

Back 22

B. RNA and proteins.

Front 23

23. 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.
E. the spliceosome.

Back 23

B. the large ribosomal subunit.

Front 24

24. In eukaryotic cells, transcription occurs
A. on the surface of the nuclear membrane.
B. on ribosomes.
C. on spliceosomes.
D. inside the nucleus.
E. on the surface of the plasma membrane.

Back 24

D. inside the nucleus.

Front 25

25. 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.
E. the sigma subunit.

Back 25

A. the holoenzyme.

Front 26

26. During _________, nucleotide sequence information is changed into amino acid sequence information.
A. replication
B. sequencing
C. transcription
D. translocation
E. translation

Back 26

E. translation

Front 27

27. The genetic code uses _________ nucleotide(s) to specify one amino acid.
A. one
B. two
C. three
D. four
E. 64

Back 27

C. three

Front 28

28. 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.
E. The genetic code is almost universal, but not quite.

Back 28

C. Every codon codes for one amino acid.

Front 29

29. How many unique mRNA codons can be constructed from the four different RNA nucleotides?
A. four
B. 12
C. 16
D. 61
E. 64

Back 29

E. 64

Front 30

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.
E. templates.

Back 30

A. codons.

Front 31

31. The tRNA nucleotide sequence that pairs with bases on the mRNA is called a(n)
A. intron.
B. exon.
C. codon.
D. initiation factor.
E. anticodon.

Back 31

E. anticodon.

Front 32

32. Ribosome movement along the mRNA is called
A. transcription.
B. initiation.
C. replication.
D. translocation.
E. activation.

Back 32

D. translocation.

Front 33

33. Specific amino acids are attached to tRNA molecules by
A. aminoacyl-tRNA synthetases.
B. hydrogen bonds.
C. anticodons.
D. deactivating enzymes.
E. initiation factors.

Back 33

A. aminoacyl-tRNA synthetases.

Front 34

34. Codons that serve as "stop" signals for translation are recognized by
A. tRNA.
B. release factors.
C. anticodons.
D. translation terminators.
E. aminoacyl-tRNA synthetases

Back 34

B. release factors.

Front 35

35. When a polypeptide is being assembled, the bond that forms between a newly added amino acid and the previous amino acid in the chain is a _________ bond.
A. hydrogen
B. hydrophobic
C. terminal
D. phosphodiester
E. peptide

Back 35

E. peptide

Front 36

36. 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.
E. initiation factors.

Back 36

D. RNA polymerase.

Front 37

37. 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.
E. noncodons.

Back 37

B. introns.

Front 38

38. The location of translation in prokaryotic cells is
A. in the nucleoid.
B. on ribosomes.
C. on the plasma membrane.
D. on mesosomes.
E. on chromosomes

Back 38

B. on ribosomes.

Front 39

39. 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.

Back 39

A. removal of exons from the pre-mRNA.

Front 40

40. During translation, uncharged tRNA molecules leave the ribosome from the _________ site.
A. E
B. P
C. A
D. termination
E. release

Back 40

A. E

Front 41

41. 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.

Back 41

D. DNA RNA proteins.

Front 42

42. 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'
E. 5' AUCGCUCC 3'

Back 42

C. 5' UAGCGAGG 3'

Front 43

43. 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.

Back 43

A. AUG CGU.

Front 44

44. 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.

Back 44

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.

Front 45

45. Eukaryotic and prokaryotic organisms differ in how they process genetic information. Which statements best explain one of these differences?
A. In prokaryotes, translation of the mRNA begins before transcription is complete. In eukaryotes, transcription and modification of the mRNA is completed before translation begins.
B. In prokaryotes, genes are transcribed directly into polypeptides. In eukaryotes, genes are transcribed into RNA which is used to assemble polypeptides.
C. In prokaryotes, translation occurs before genes are transcribed into mRNA. In eukaryotes, genes are transcribed into mRNA which is then translated into polypeptides.
D. In prokaryotes, introns are removed before genes are transcribed into mRNA. In eukaryotes, introns are removed after genes are transcribed into mRNA.

Back 45

A. In prokaryotes, translation of the mRNA begins before transcription is complete. In eukaryotes, transcription and modification of the mRNA is completed before translation begins.

Front 46

46. 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
E. transcription factors bind to the TATA box sequence

Back 46

B. RNA polymerase binds to the promoter

Front 47

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.

Back 47

C. a primer.

Front 48

48. Transcription in prokaryotes is carried out by ______, which unwind(s) and transcribe(s) the gene.
A. RNA synthetase
B. RNA polymerase II
C. RNA polymerase III
D. transcription factors
E. RNA polymerase

Back 48

E. RNA polymerase

Front 49

49. Eukaryotes have ____ type(s) of RNA polymerase.
A. two
B. three
C. four
D. one
E. 64

Back 49

B. three

Front 50

50. In eukaryotes, each type of RNA polymerase recognizes a different
A. start codon
B. stop codon
C. promoter
D. release factor
E. transcription factor

Back 50

C. promoter

Front 51

51. 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.
E. at the transcription bubble.

Back 51

C. inside the nucleus.

Front 52

52. In eukaryotes, the 3' poly-A tail is attached to
A. poly-A polymerase.
B. mRNA.
C. tRNA.
D. the ribosome.
E. the template strand of DNA.

Back 52

B. mRNA.

Front 53

53. 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.

Back 53

C. There is some flexibility in pairing between the 3' base of the codon and the 5' base of the anticodon.

Front 54

54. During translation, translocation refers to
A. releasing a tRNA molecule from the ribosome.
B. joining an amino acid to a tRNA molecule.
C. joining an amino acid to the next amino acid in the chain.
D. joining a tRNA molecule to the ribosome.
E. moving the ribosome along the mRNA molecule.

Back 54

E. moving the ribosome along the mRNA molecule.

Front 55

55. 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.
E. The product of transcription, called the primary transcript, is cut and all pieces are joined back together to form the mature mRNA.

Back 55

D. The product of transcription, called the primary transcript, is cut and some pieces are joined back together to form the mature mRNA.

Front 56

56. 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
E. tRNA

Back 56

C. mRNA

Front 57

57. Which molecule combines with proteins to form both the large and small ribosomal subunits?
A. DNA
B. RNA polymerase
C. miRNA
D. rRNA
E. tRNA

Back 57

D. rRNA

Front 58

58. You are conducting a genetic screen to isolate nutritional mutants in yeast. Specifically, you want to isolate a double mutant that cannot synthesize histidine or leucine, two nutrients essential for growth. You start with a wild type yeast strain and mutagenize it with a UV light. Which of the following outlines the remaining steps for isolating such a mutant?
A. Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine.
B. Grow mutagenized yeast on minimal media. Then grow them on rich media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on rich media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine.
C. Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that grow on minimal media, grow on media supplemented with only histidine or leucine, but do not grow on minimal media supplemented with both histidine and leucine.
D. Grow mutagenized yeast on minimal media. Then grow them on rich media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that grow on rich media, grow on media supplemented with only histidine or leucine, but do not grow on minimal media supplemented with both histidine and leucine.

Back 58

A. Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine.

Front 59

59. 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 THE FAT CAT ATE THE RED RAT

Back 59

C. THE FAC ATA TET HER EDR AT

Front 60

60. What would happen if snRNAs did not recognize the branch point within an intron?
A. A lariat would not form.
B. snRNAs 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.

Back 60

A. A lariat would not form.

Front 61

61. During the splicing reaction, the intron-exon junctions are recognized by
A. snRNPs.
B. miRNAs.
C. SRP RNAs.
D. the lariat.
E. the branch point.

Back 61

A. snRNPs.

Front 62

62. 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.
E. two alpha subunits, two beta subunits, and two sigma subunits.

Back 62

D. the core polymerase plus a sigma subunit.

Front 63

63. 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.
E. It allows RNA polymerase to distinguish between the template strand and the coding strand of the DNA molecule.

Back 63

B. Both the location of the start site and the direction of transcription can be established.

Front 64

64. 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.
E. The position of the 5' end of the RNA would be unstable, stimulating elongation.

Back 64

D. The position of the 3' end of the RNA would be unstable, inhibiting elongation.

Front 65

65. 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

Back 65

A. block transcription

Front 66

66. 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
E. binding of the sigma subunit to promoter elements at -35 and -10, followed by recruitment of the core polymerase

Back 66

A. binding of a transcription factor to the TATA box, followed by recruitment of additional transcription factors and recruitment of RNA polymerase II

Front 67

67. 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
E. failure of snRNPs to combine with protein and form the spliceosome

Back 67

B. failure to form a lariat

Front 68

68. 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. There is no modification of the primary transcript in the pancreas.
C. There is no modification of the primary transcript in the brain, muscle, and liver.
D. Alternative splicing in the pancreas yields a protein that is missing the portion that the antibody recognizes.
E. Alternative splicing in the brain, muscle, and liver increases the level of translation.

Back 68

D. Alternative splicing in the pancreas yields a protein that is missing the portion that the antibody recognizes.

Front 69

69. The mutation responsible for Huntington's disease is a
A. missense mutation.
B. nonsense mutation.
C. frameshift mutation.
D. triplet repeat expansion mutation.

Back 69

D. triplet repeat expansion mutation.

Front 70

70. 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.

Back 70

A. The inversion would probably have no effect on gene expression.

Front 71

71. 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.

Back 71

D. Only DNA polymerase has a proofreading ability.

Front 72

72. 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
E. AAA codes for phenylalanine and AUA codes for isoleucine

Back 72

C. AAA codes for lysine

Front 73

73. 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.
E. translation termination.

Back 73

A. transcription initiation.

Front 74

74. What is the best way to describe our current understanding of the one-gene/one-polypeptide hypothesis?
A. It applies to both prokaryotes and eukaryotes.
B. It applies to prokaryotes but not to eukaryotes.
C. It applies to eukaryotes but not to prokaryotes.
D. It has been replaced by the one-gene/one-enzyme hypothesis.

Back 74

B. It applies to prokaryotes but not to eukaryotes.

Front 75

75. 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

Back 75

D. RNA polymerase I

Front 76

76. 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 most of it is accumulating in the cytoplasm rather than being secreted from the cells. Which hypothesis to explain this observation makes the most sense?
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.
E. A missense mutation has altered the ribosome-binding sequence at the 5' end of the mRNA.

Back 76

A. A small deletion has removed the nucleotides that code for the signal sequence at the amino terminus of the protein.

Front 77

77. 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.

Back 77

C. Eukaryotic genes often contain introns while prokaryotic genes do not.

Front 78

78. What is the base sequence, in the DNA template strand, of the intron that is closest to the 3' end of this strand?

Shown below is a hypothetical DNA sequence from a virus. Also shown is the sequence of the RNA that is synthesized from this DNA.

DNA sequence:

5'-AGCACCTGCCGAATGGGCCAAATCCTGCCGAATAAA-3'
3'-TCGTGGACGGCTTACCCGGTTTAGGACGGCTTATTT -5'

RNA sequence (G* = G cap):

5'-G*AGCACCUGCCGCCUGCCGAAUAAAAAAA....-3'

78. What is the base sequence, in the DNA template strand, of the intron that is closest to the 3' end of this strand?
A. TCGTGGACGGC
B. TTACCCGGTTTA
C. GGACGGCTTATTT
D. GCTTACCCGGTT

A. TCGTGGACGGC
B. TTACCCGGTTTA
C. GGACGGCTTATTT
D. GCTTACCCGGTT

Back 78

B. TTACCCGGTTTA

Front 79

79. The RNA was most likely transcribed by
A. RNA polymerase holoenzyme.
B. RNA polymerase I.
C. RNA polymerase II.
D. RNA polymerase III.

Back 79

B. RNA polymerase I.