All EXCEPT which of the following are characteristics of the genetic
material?
A. It must be replicated accurately.
B. It must be capable
of change.
C. It contains all the information needed for growth,
development, and reproduction of the organism.
D. It is composed
of protein.
D. It is composed of protein.
(Although early observations favored protein as the genetic
material, subsequent experiments demonstrated that the genetic
material was nucleic acid.)
The Hershey and Chase experiments involved the preparation of two
different types of radioactively labeled phage. Which of the following
best explains why two preparations were required?
A. It was necessary that each of the two phage components, DNA
and protein, be identifiable upon recovery at the end of the
experiment.
B. Each scientist had his own method for labeling
phage, so each conducted the same experiment using a different
isotope.
C. The bacteriophage used in the experiments was a T2
phage.
D. Establishing the identity of the genetic material
required observation of two phage generations.
A. It was necessary that each of the two phage components, DNA and
protein, be identifiable upon recovery at the end of the experiment.
(Because it was concluded that the component associated with
bacteria at the end of the experiment must be the genetic material, it
was critical that the component be identifiable as either DNA or protein.)
Which of the following statements best represents the central
conclusion of the Hershey-Chase experiments?
A. When radioactive sulfur is supplied in a growth medium, it is
primarily DNA that incorporates radioactive label.
B. Phage T2
is capable of replicating within a bacterial host.
C. DNA is the
identity of the hereditary material in phage T2.
D. Some viruses
can infect bacteria.
C. DNA is the identity of the hereditary material in phage T2.
(Because phage DNA and not protein was associated with bacteria
at the end of the experiment, it could be concluded that DNA - not
protein - must be the genetic material.)
Which of the following outcomes would be most likely if the
Hershey-Chase experiments were repeated without the step involving the
blender?
A. Neither preparation of infected bacteria would exhibit
radioactivity.
B. Both preparations of infected bacteria would
contain both P32 and S35.
C. Both preparations of infected
bacteria would exhibit radioactivity.
D. The phage would fail to
infect bacteria.
C. Both preparations of infected bacteria would exhibit
radioactivity.
(Instead of being removed from the preparation, the
"ghosts" would be retained. Because both bacterial
preparations would include ghosts as well as viral DNA, both would be
radioactive, one with P32, one with S35.)
What observation did Griffith make in his experiments with
Streptococcus pneumoniae?
A. The heat-killed, virulent Streptococcus pneumoniae was lethal
to the mouse.
B. That DNA is the genetic material.
C. The
mouse survived injection of live virulent (smooth) Streptococcus
pneumoniae.
D. The mouse did not survive when injected with a
mixture of live, avirulent (rough) Streptococcus pneumoniae and
heat-killed, virulent Streptococcus pneumoniae.
D. The mouse did not survive when injected with a mixture of live,
avirulent (rough) Streptococcus pneumoniae and heat-killed, virulent
Streptococcus pneumoniae.
(Something in the heat-killed preparation was able to transform
the avirulent strain to a virulent form.)
What results did Avery, McLeod, and McCarty obtain in their
experiments with virulent bacteria?
A. DNase destroyed the transforming activity.
B. Protease
destroyed the transforming activity.
C. RNase destroyed the
transforming activity.
D. The transforming principle was too
complex and difficult to be purified.
A. DNase destroyed the transforming activity.
(Treatment of the transforming principle with DNase destroyed
the DNA and thus its ability to transform bacteria.)
Guanine and adenine are purines found in DNA.
A. True
B. False
A. True
(Guanine and adenine are indeed purines found in DNA; thymine
and cytosine are the pyrimidines found in DNA.)
Which of the following statements about DNA structure is true?
A. The nucleic acid strands in a DNA molecule are oriented
antiparallel to each other, meaning they run in opposite directions.
B. Nucleic acids are formed through phosphodiester bonds that
link nucleosides together.
C. Hydrogen bonds formed between the
sugar‑phosphate backbones of the two DNA chains help to stabilize DNA
structure.
D. The pentose sugar in DNA is ribose.
A. The nucleic acid strands in a DNA molecule are oriented
antiparallel to each other, meaning they run in opposite directions.
(This statement is true; the 5′–3′ orientation of each chain
runs in opposite directions.)
What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′?
A. 5′ CAGTCAAGCAT 3′
B. 5′ ATGCTTGACTG 3′
C. 5′
ACTCTACGTAG 3′
D. 5′ TACGAACTGAC 3′
A. 5′ CAGTCAAGCAT 3′
(This sequence is complementary and in the correct orientation.)
The results of the Meselson-Stahl experiments relied on all of the
following except _______.
A. that a heavy isotope of nitrogen could be incorporated into
replicating DNA molecules
B. a cesium chloride gradient
C.
the fact that DNA is the genetic material
D. a means of
distinguishing among the distribution patterns of newly synthesized
and parent molecule DNA possible
C. the fact that DNA is the genetic material
(Correct. This fact had already been established and was not of
any consequence in these experiments.)
After observing the results of one round of replication, the
scientists obtained results from a second round. The purpose of one
additional round of replication was to _______.
A.distinguish between semi-conservative and dispersive
replication
B. distinguish between conservative and dispersive
replication
C. confirm that replication is conservative
D.
distinguish between conservative and semi-conservative replication
A. distinguish between semi-conservative and dispersive replication
The enzyme that can replicate DNA is called?
DNA polymerase
Which of the following would result from a third round of replication
using the methods of Meselson and Stahl?
A. One heavy band, one light band, and one intermediate band
B. One light band and one intermediate band
C. One light
band
D. One heavy band
B. One light band and one intermediate band
(Correct. Of the molecules generated in the third round, 75% are
completely light, 25% are intermediate.)
In the Meselson-Stahl experiment, which mode of replication was
eliminated based on data derived after one generation of replication?
A. conservative
B. semiconservative
C. dispersive
D. none of the modes
A. conservative
(The conservative replication theory says that parental strands
reanneal with parental strands, and daughter strands reanneal with
daughter strands after DNA replication. This experiment showed that
this was not the case.)
The new DNA strand that grows continuously in the 5' to 3' direction is called the
leading strand
During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA is called a
replication fork
After replication is complete, the new DNAs called ____, are identical to each other
daughter DNA
_____ are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA
Okazaki fragments
Leading Strand
Only one primer needed, made continuously, and daughter strand
elongates toward replication fork.
The leading strand is made continuously from a single RNA primer
located at the origin of replication. DNA pol III adds nucleotides to
the 3' end of the leading strand so that it elongates toward the
replication fork.
Lagging Strand
Multiple primers needed, daughter strand elongates away from
replication fork, and made in segments.
In contrast, the lagging strand is made in segments, each with
its own RNA primer. DNA pol III adds nucleotides to the 3' end of the
lagging strand so that it elongates away from the replication fork.
Both leading and lagging strand _____
Synthesized 5' to 3'
The leading and lagging strands built on the same template
strand will eventually be joined, forming a continuous daughter strand.
Helicase
binds at the replication fork and breaks H-bonds between bases
Topoisomerase
binds ahead of the replication fork and breaks covalent bonds in DNA backbone
Single-Strand Binding Protein
Binds after the replication fork and prevents H-bonds between bases