Download Chapter 10 retake assignment with KEY

Ch.10 Test Retake Assignment- Answer on a separate sheet of paper
Short Answer
Explain how the bacteria in Griffith’s experiment were transformed.
What is a bacteriophage?
What causes translation to stop?
What is the template mechanism of DNA replication?
What is the “one gene–one polypeptide” hypothesis?
What are the three main parts of an RNA nucleotide?
If codons consisted of fewer than 3 bases, could they still code for 20 amino acids?
What must happen to a DNA molecule before RNA polymerase can make RNA?
If the percentage of guanine in the DNA of a certain species decreased by 5 percent over time, what
would you expect to have happened to the percentage of adenine in that DNA?
52. What is a mutation?
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Figure 11-1
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Observing How are structures A and B in Figure 11-1 related?
Observing What is structure E in Figure 11-1? What does it code for?
Predicting What would happen to structure F in Figure 11-1 if structure C were deleted?
Observing Identify structure F in Figure 11-1. What does it specify?
Inferring From which labeled structure in Figure 11-1 is structure D made? Identify that labeled
structure.
58. Predicting In Figure 11-1, what effect would the deletion of structure C have on the process that
occurs during step Y?
Figure 11-2
59. Observing Identify structure X in Figure 11-2.
60. Observing What molecule’s structure is illustrated in Figure 11-2?
61. Inferring What are the dashed lines shown in Figure 11-2? How do they relate to the rules for base
pairing?
62. Observing What is the twisting shape of the molecule in Figure 11-2 called?
63. Observing What is a pyrimidine? Identify two pyrimidines in Figure 11-2.
64. Observing What is a purine? Identify two purines in Figure 11-2.
Essay
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Describe the process in which an RNA transcript is converted into a final mRNA.
Contrast the functions of the three main types of RNA.
Why do some kinds of mutations cause greater changes in proteins than others?
Describe some causes of mutations. Use the term mutagen in your answer.
Describe the Hershey-Chase experiment. Why were the results important?
Describe the structure of a DNA molecule.
Describe the process of DNA replication in eukaryotes.
How does transcription differ from DNA replication? Describe at least four differences.
SHORT ANSWER
43. ANS:
The harmless living bacteria took in pneumonia-causing DNA (genes) from the heat-killed, pneumonia-causing
bacteria, as a result of which the harmless bacteria changed into bacteria that cause pneumonia.
NOT: core
44. ANS:
A bacteriophage is a kind of virus that infects bacteria.
NOT: core
45. ANS:
A stop codon on the mRNA causes translation to stop.
NOT: core
46. ANS:
The base-pairing rules mean that if you know the sequence of bases on one strand of DNA, you can determine
the sequence on the other strand. During DNA copying, the two strands of DNA separate and each strand acts as
a template for producing a new, complementary strand.
NOT: core
47. ANS:
The “one gene–one polypeptide” hypothesis is the idea that each gene dictates the production of a single
polypeptide, which may make up part of an enzyme or another kind of protein.
NOT: core
48. ANS:
A ribose molecule, a phosphate group, and a nitrogenous base are the three main parts of an RNA nucleotide.
49. ANS:
If codons consisted of fewer than 3 bases, the bases could not combine in enough ways to produce codons for
specifying the 20 different amino acids.
50. ANS:
The DNA molecule must be separated into two strands.
51. ANS:
The percentage of adenine would have increased by about 5 percent.
52. ANS:
A mutation is any change in the nucleotide sequence of DNA.
OTHER
53. ANS:
Structures A and B are complementary strands of DNA.
NOT: core
54. ANS:
Structure E is the start codon, which codes the amino acid methionine.
NOT: core
55. ANS:
The base sequence of the codon (structure F) would change from GCU to GUG.
NOT: core
56. ANS:
Structure F is a codon that specifies the amino acid alanine.
NOT: core
57. ANS:
Structure D is made from structure A, which is one of the strands of DNA.
NOT: core
58. ANS:
The deletion of structure C would alter the codons during translation. As a result, the sequence of the amino
acids that make up the growing polypeptide would change.
NOT: core
59. ANS:
Structure X is a nucleotide (cytosine).
60. ANS:
Figure 11-2 shows a DNA molecule.
61. ANS:
The dashed lines are hydrogen bonds. Pairings between nucleotides are partly determined by the ability of
nucleotides to form hydrogen bonds with each other.
62. ANS:
The shape is called a double helix.
63. ANS:
A pyrimidine is a single-ringed nitrogenous base. The structures labeled C and T are pyrimidines.
64. ANS:
A purine is a double-ringed nitrogenous base. The structures labeled A and G are pyrimidines.
ESSAY
65. ANS:
After being transcribed from DNA, the RNA transcript is reduced in size by having noncoding regions, called
introns, cut out from it. The remaining coding regions, called exons, are then spliced together.
NOT: core
66. ANS:
Messenger RNA carries copies of instructions for assembling proteins from DNA to the ribosomes. Ribosomal
RNA is a component of the ribosomes. Transfer RNA translates the codons of mRNA into the amino acids that
make up proteins.
NOT: core
67. ANS:
Mutations include base substitutions, base insertions, and base deletions of single nucleotides in DNA.
Insertions and deletions have a greater effect on proteins than do substitutions, because insertions and deletions
affect every amino acid that is specified by the nucleotides that follow the point of mutation. In contrast, a
substitution affects a single amino acid. A change in more than one amino acid is more likely to alter the ability
of the protein to function normally than is a change in a single amino acid.
NOT: core
68. ANS:
Mutations can be caused by chemical or physical agents, called mutagens. One physical mutagen is high-energy
radiation, such as X-rays and ultraviolet light. One type of chemical mutagen consists of chemicals that are
similar to normal DNA bases but cause incorrect base-pairing when incorporated into DNA.
NOT: core
69. ANS:
Hershey and Chase grew bacteriophages in cultures containing radioactive isotopes of phosphorus and sulfur.
The radioactive phosphorus became incorporated into the bacteriophage’s DNA, because DNA contains
phosphorus. The radioactive sulfur became incorporated into the bacteriophage’s protein coat, because proteins
contain sulfur. After the bacteriophages were allowed to infect bacteria, Hershey and Chase found that nearly all
the radioactivity in the bacteria was from phosphorus. This indicated that the bacteriophage’s DNA was injected
into the bacteria. The results were important because they showed that the bacteriophage’s genetic material was
DNA, not protein.
70. ANS:
A DNA molecule has the shape of a double helix, or that of a twisted ladder. Each strand of the helix is a chain
of nucleotides. The two strands are held together by hydrogen bonds between the nitrogenous bases of the
nucleotides on opposite strands. The nitrogenous bases form hydrogen bonds with one another in pairs. Adenine
forms hydrogen bonds with thymine, and guanine forms hydrogen bonds with cytosine.
71. ANS:
DNA replication begins with the two DNA strands separating and unwinding. Each strand serves as a template
for the attachment of complementary bases, forming a new strand of DNA. As a result, two identical DNA
molecules are formed, each with one original strand of DNA and one new strand of DNA. DNA replication
begins at hundreds of points on the chromosomes (origins of replication) and proceeds in two directions away
from these points.
72. ANS:
RNA polymerase is involved in transcription, whereas DNA polymerase is involved in DNA replication. During
transcription, free nucleotides base pair with the nucleotides on only one strand of a DNA molecule, not both
strands as in DNA replication. In transcription, the free nucleotides are RNA nucleotides, not DNA nucleotides.
Transcription continues until a stop signal is reached on the DNA strand. DNA replication continues until the
entire chromosome is replicated. At the end of transcription, one single-stranded RNA molecule is formed, not
two double-stranded DNA molecules. The newly formed RNA molecule leaves the nucleus, whereas the newly
formed DNA molecules stay in the nucleus.