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Raven/Johnson Biology 8e
Chapter 14 - Answers
1. What was the key finding from Griffith’s experiments using live and heat-killed
pathogenic bacteria?
a. Bacteria with a smooth coat could kill mice.
b. Bacteria with a rough coat are not lethal.
c. Heat-killed smooth-coat bacteria would not cause death in mice.
d. Heat-killed smooth-coat bacteria could transform the nonlethal live bacteria.
The correct answer is d—
A. Answer a is incorrect. The lethality of smooth-coat bacteria was known before Griffith
started the experiment.
The correct answer is d—
B. Answer b is incorrect. This observation was not a result of this experiment.
The correct answer is d—
C. Answer c is incorrect. This is a result from this experiment; however, it is not the
important result.
The correct answer is d—Heat-killed smooth-coat bacteria could transform the nonlethal live
bacteria.
D. Answer d is correct. Transformation of nonlethal cells to lethal cells was the critical result
that demonstrated the ability of genetic material to be transferred between cells.
2. When Hershey and Chase differentially tagged the DNA and proteins of bacteriophages
and allowed them to infect bacteria, what did the viruses transfer to the bacteria?
a. Radioactive phosphorous and sulfur
b. Radioactive sulfur
c. DNA
d. Both b and c
The correct answer is c—
A. Answer a is incorrect. Radioactive phosphorous labeled the DNA and radioactive sulfur
labeled protein. If both phosphorous and sulfur were transferred to the bacteria that would
mean that both DNA and protein are transferred from the virus.
The correct answer is c—
B. Answer b is incorrect. Radioactive sulfur labeled proteins. Proteins were not transferred to
the bacteria.
The correct answer is c—DNA
C. Answer c is correct. The experiment demonstrated that DNA is transferred from the virus
to the bacteria.
The correct answer is c—
D. Answer d is incorrect. Radioactive sulfur labeled proteins. Proteins were not transferred to
the bacteria.
3. Which of the following is NOT a component of DNA?
a. Pyrimidines such as uracil
Raven/Johnson Biology 8e
Chapter 14 - Answers
b. Five-carbon sugars
c. Purines such as adenine
d. Phosphate groups
The correct answer is a—Pyrimidines such as uracil
A. Answer a is correct. Uracil is a pyrimidine that is found in RNA.
The correct answer is a—
B. Answer b is incorrect. DNA is composed of ribose—a five-carbon sugar.
The correct answer is a—
C. Answer c is incorrect. The purine adenine is present in DNA.
The correct answer is a—
D. Answer d is incorrect. All the nucleotides in DNA are composed of a phosphate group.
4. What type of chemical bond allows DNA or RNA to form a long polymer?
a. Hydrogen bonds
b. Peptide bonds
c. Ionic bonds
d. Phosphodiester bonds
The correct answer is d—
A. Answer a is incorrect. Hydrogen bonds are important for the formation complementary
base pairing.
The correct answer is d—
B. Answer b is incorrect. Peptide bonds are involved in forming proteins.
The correct answer is d—
C. Answer c is incorrect. Ionic bonds are not involved in the formation of nucleic acid
polymers.
The correct answer is d—Phosphodiester bonds
D. Answer d is correct. Phosphodiester bonds are the covalent bonds formed between the
sugar and phosphate groups of nucleic acids.
5. What is Chargaff’s rule?
a. The number of phosphate groups always equals the number of five-carbon sugars.
b. The proportions of A equal that of C and G equals T.
c. The proportions of A equal that of T and G equals C.
d. Purines binds to pyrimidines
The correct answer is c—
A. Answer a is incorrect. Chargaff’s rule refers to the numbers of purines and pyrimidines.
Since every nucleic acid is composed of a sugar and phosphate group there would be no
variation.
The correct answer is c—
Raven/Johnson Biology 8e
Chapter 14 - Answers
B. Answer b is incorrect. Chargaff’s rule is based on complementary base-pairing. Adenine
does not base pair with cytosine.
The correct answer is c—The proportions of A equal that of T and G equals C.
C. Answer c is correct. Chargaff’s rule is based on complementary base-pairing. Adenine
base-pairs with thymine, and guanine base-pairs with cytosine.
The correct answer is c—
D. Answer d is incorrect. Chargaff’s rule is related to the relative numbers of purines and
pyrimidines. It does not address the ability of the nucleotides to form hydrogen bonds.
6. The bonds that hold two complementary strands of DNA together are—
a. Hydrogen bonds
b. Peptide bonds
c. Ionic bonds
d. Phosphodiester bonds
The correct answer is a—Hydrogen bonds
A. Answer a is correct. Hydrogen bonds are formed between the nitrogenous bases of the
nucleotides.
The correct answer is a—
B. Answer b is incorrect. Peptide bonds are found in proteins, not nucleic acids.
The correct answer is a—
C. Answer c is incorrect. Ionic bonds are not found in nucleic acids.
The correct answer is a—
D. Answer d is incorrect. Phosphodiester bonds make up the sugar–phosphate backbone of
the nucleic acid polymer.
7. If one strand of a DNA molecule has the sequence ATTGCAT, then the complementary
strand will have the sequence—
a. ATTGCAT
b. TACGTTA
c. TAACGTA
d. GCCTAGC
The correct answer is c—
A. Answer a is incorrect. This is an identical sequence. Adenine will not form a hydrogen
bond with itself.
The correct answer is c—
B. Answer b is incorrect. This cannot be the complementary sequence since the base-pairing
is not correct. A binds to T and G binds to C.
The correct answer is c—TAACGTA
C. Answer c is correct. The base-pairing rule is that A binds to T and G binds to C. This
sequence meets this rule.
Raven/Johnson Biology 8e
Chapter 14 - Answers
The correct answer is c—
D. Answer d is incorrect. The base-pairing rule is that A binds to T and G binds to C. This
sequence does not meet this rule.
8. Which of the following is NOT part of the Watson–Crick model of the structure of DNA?
a. DNA is composed of two strands.
b. The two DNA strands are oriented in parallel (5′ to 3′).
c. Purines bind to pyrimidines.
d. DNA forms a double helix.
The correct answer is b—
A. Answer a is incorrect. The Watson–Crick model proposes that DNA is composed of two
strands.
The correct answer is b—The two DNA strands are oriented in parallel (5′ to 3′).
B. Answer b is correct. The Watson–Crick model proposes that DNA is composed of
antiparallel strands—one running 5′ to 3′ and the other 3′ to 5′.
The correct answer is b—
C. Answer c is incorrect. Base-pairing between purine and pyrimidines is a part of the
Watson–Crick model of DNA.
The correct answer is b—
D. Answer d is incorrect. The double-helix structure of DNA is a central feature of the
Watson–Crick model.
9. Meselson and Stahl demonstrated that—
a. DNA replication occurs in bacteria
b. DNA replication is dispersive
c. DNA replication is conservative
d. DNA replication is semiconservative
The correct answer is d—
A. Answer a is incorrect. Meselson and Stahl examined the mechanism of DNA replication,
not what type of cells can replicate DNA.
The correct answer is d—
B. Answer b is incorrect. Dispersive replication would lead to the formation of strands of
DNA that were a mix of parental and new DNA. This did not occur.
The correct answer is d—
C. Answer c is incorrect. Conservative replication would lead to the formation of separate
strands of parental and new DNA. This did not occur.
The correct answer is d—DNA replication is semiconservative
D. Answer d is correct. The newly synthesized DNA is a combination of one parental strand
and one new strand.
Raven/Johnson Biology 8e
Chapter 14 - Answers
10. Which of the following steps in DNA replication involves the formation of new
phosphodiester bonds?
a. Initiation at an origin of replication
b. Elongation by a DNA polymerase
c. Unwinding of the double helix
d. Termination
The correct answer is b—
A. Answer a is incorrect. Initiation involves the opening up of the DNA helix.
The correct answer is b—Elongation by a DNA polymerase
B. Answer b is correct. Elongation involves the formation of phosphodiester bonds between
the new nucleotides through the activity of DNA polymerases.
The correct answer is b—
C. Answer c is incorrect. Unwinding of the helix is important for replication; however, this
step does not involve bond formation.
The correct answer is b—
D. Answer d is incorrect. Termination is the point where the formation of new
phosphodiester bond formation stops.
11. The difference in leading- versus lagging-strand synthesis is a consequence of—
a. the antiparallel configuration of DNA
b. DNA polymerase III synthesizing only 5′ to 3′
c. the activity of DNA gyrase
d. both a and b
The correct answer is d—
A. Answer a is incorrect. The antiparallel configuration contributes to the difference between
leading- and lagging-strand synthesis, but it is not the only factor.
The correct answer is d—
B. Answer b is incorrect. The 5′ to 3′ activity (true of all DNA polymerases) of DNA
polymerase III contributes to the differences between leading- and lagging-strand
synthesis, but is not the only factor.
The correct answer is d—
C. Answer c is incorrect. The activity of DNA gyrase does not contribute to the difference
between leading- and lagging-strand synthesis.
The correct answer is d—both a and b
D. Answer d is correct. DNA polymerase III synthesizes DNA only in a 5′ to 3′ direction;
however, because DNA has an antiparallel orientation there is a difference in the rate of
synthesis between the two strands.
12. Okazaki fragments are—
a. synthesized in the 3′ to 5′ direction
b. found on the lagging strand
Raven/Johnson Biology 8e
Chapter 14 - Answers
c. found on the leading strand
d. made of RNA
The correct answer is b—
A. Answer a is incorrect. An Okazaki fragment is a segment of DNA and DNA is
synthesized in the 5′ to 3′ direction.
The correct answer is b—found on the lagging strand
B. Answer b is correct. Okazaki fragments are only found in the lagging strand as a
consequence of discontinuous synthesis.
The correct answer is b—
C. Answer c is incorrect. The leading strand undergoes continuous synthesis and therefore
does not generate Okazaki fragments.
The correct answer is b—
D. Answer d is incorrect. Okazaki fragments are made of DNA. The primers are composed
of RNA.
13. Successful DNA synthesis requires all of the following except—
a. helicase
b. endonuclease
c. DNA primase
d. DNA ligase
The correct answer is b—
A. Answer a is incorrect. The helicase is required to unwind the DNA, making it possible for
the polymerase to bind to the replication fork.
The correct answer is b—endonuclease
B. Answer b is correct. Endonuclease function is important for DNA repair. DNA
polymerases contain exonuclease activity.
The correct answer is b—
C. Answer c is incorrect. Primase generates the RNA primer required to start DNA
polymerase III synthesis.
The correct answer is b—
D. Answer d is incorrect. Ligase activity is required to generate phosphodiester bonds to link
Okazaki fragments.
14. What is a telomere?
a. An A-T rich region of DNA
b. The point of DNA termination on a bacterial chromosome
c. Regions of repeated sequences of DNA on the ends of eukaryotic chromosomes
d. The sequence of RNA found on a replicating molecule of DNA
The correct answer is c—
Raven/Johnson Biology 8e
Chapter 14 - Answers
A. Answer a is incorrect. An origin of replication is characterized by the high proportion of
adenine and thymine.
The correct answer is c—
B. Answer b is incorrect. Bacteria possess circular chromosomes. They do not have
telomeres.
The correct answer is c—Regions of repeated sequences of DNA on the ends of eukaryotic
chromosomes
C. Answer c is correct. Telomeres are found at the end of eukaryotic chromosomes. They are
characterized by repeated sequences of DNA.
The correct answer is c—
D. Answer d is incorrect. The sequence of RNA used in DNA replication is known as a
primer.
15. Which type of enzyme is involved in excision repair?
a. photolyase
b. DNA polymerase III
c. endonuclease
d. telomerase
The correct answer is c—
A. Answer a is incorrect. A photolyase is the enzyme involved in repair of a thymine–
thymine dimer.
The correct answer is c—
B. Answer b is incorrect. DNA polymerase III is involved in the normal synthesis of
DNA.
The correct answer is c—endonuclease
C. Answer c is correct. An endonuclease can break the phosphodiester bonds between the
nucleotides in a damaged region of DNA.
The correct answer is c—
D. Answer d is incorrect. A telomerase is the enzyme that catalyzes the synthesis of
DNA at the ends of eukaryotic chromosomes.
Challenge Questions
1. The work by Griffith provided the first indication that DNA was the genetic material.
Review the four experiments outline in Figure 14.1. Predict the likely outcome for the
following variations on this classic research.
a. Heat-killed pathogenic and heat-killed nonpathogenic
b. Heat-killed pathogenic and live nonpathogenic in the presence of an enzyme
that digests proteins (proteases)
c. Heat-killed pathogenic and live nonpathogenic in the presence of an enzyme
that digests DNA (endonuclease)
Raven/Johnson Biology 8e
Chapter 14 - Answers
Answer—The key finding from the Griffith experiments is that heat-killed pathogenic bacteria
can “transform” living, nonpathogenic bacteria. It was later discovered that the transforming
factor was DNA.
a. If both bacteria are heat-killed, then the transfer of DNA will have no effect since
pathogenicity requires the production of proteins encoded by the DNA. Protein
synthesis will not occur in a dead cell.
b. The nonpathogenic cells will be transformed to pathogenic cells. Loss of proteins will
not alter DNA.
c. The nonpathogenic cells remain nonpathogenic. If the DNA is digested, it will not be
transferred and no transformation will occur.
2. Imagine that you have identified the sequence 5′-TTATAAAGCAATAGT-3′ in a
eukaryotic chromosome. Could this region of the chromosome function as an origin of
replication? Predict the sequence of an RNA primer that would be formed in
association with this sequence.
Answer—The region could be an origin of replication. Origins of replication are adenine- and
thymine-rich regions since only these nucleotides form two hydrogen bonds versus the three
hydrogen bonds formed between guanine and cytosine, making it easier to separate the two
strands of DNA.
The RNA primer sequences would be 5′-ACUAUUGCUUUAUAA-3′. The sequence is
antiparallel to the DNA sequence (review Figure 14.16) meaning that the 5′ end of the RNA is
matching up with the 3′ end of the DNA. It is also important to remember that in RNA the
thymine nucleotide is replaced by uracil (U). Therefore, the adenine in DNA will form a
complementary base-pair with uracil.
3. Enzyme function is critically important for the proper replication of DNA. Predict the
consequence of a loss of function for each of the following enzymes.
a. DNA gyrase
b. DNA polymerase III
c. DNA ligase
d. DNA polymerase I
Answer—Consider the role of each enzyme in DNA replication to get an insight into what a loss
of function might mean for the cell.
a. DNA gyrase functions to relieve torsional strain on the DNA. If DNA gyrase were not
functioning, the DNA molecule would undergo supercoiling, causing the DNA to
wind up on itself, preventing the continued binding of the polymerases necessary for
replication.
b. DNA polymerase III is the primary polymerase involved in the addition of new
nucleotides to the growing polymer and in the formation of the phosphodiester bonds
that make up the sugar–phosphate backbone. If this enzyme were not functioning,
then no new DNA strand would be synthesized and there would be no replication.
c. DNA ligase is involved in the formation of phosphodiester bonds between Okazaki
fragments. If this enzyme was not functioning, then the fragments would remain
disconnected and would be more susceptible to digestion by nucleases.
d. DNA polymerase I functions to remove and replace the RNA primers that are required
for DNA polymerase III function. If DNA polymerase I was not available, then the
Raven/Johnson Biology 8e
Chapter 14 - Answers
RNA primers would remain and the replicated DNA would become a mix of DNA
and RNA.