BIO 340 - Exam 3 Quizlet Questions And
Answers With Verified Solutions Graded A+ Latest
Update 2025.
All EXCEPT which of the following are characteristics of the genetic material?
A. It must be replicated accurately.
B. It contains all the information needed for growth, development, and reproduction of the organism.
C. It must be capable of change.
D. It is composed of protein. - ANSWER 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. Each scientist had his own method for labeling phage, so each conducted the same experiment using a
different isotope.
B. Establishing the identity of the genetic material required observation of two phage generations.
C. The bacteriophage used in the experiments was a T2 phage.
D. It was necessary that each of the two phage components, DNA and protein, be identifiable upon
recovery at the end of the experiment. - ANSWER D. 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. Phage T2 is capable of replicating within a bacterial host.
B. DNA is the identity of the hereditary material in phage T2.
C. Some viruses can infect bacteria.
,D. When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates
radioactive label. - ANSWER B. 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. The phage would fail to infect bacteria.
B. Neither preparation of infected bacteria would exhibit radioactivity.
C. Both preparations of infected bacteria would exhibit radioactivity.
D. Both preparations of infected bacteria would contain both P32 and S35. - ANSWER 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. That DNA is the genetic material.
B. The heat-killed, virulent Streptococcus pneumoniae was lethal to the mouse.
C. 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 survived injection of live virulent (smooth) Streptococcus pneumoniae. - ANSWER
C. 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. Protease destroyed the transforming activity.
B. The transforming principle was too complex and difficult to be purified.
C. RNase destroyed the transforming activity.
D. DNase destroyed the transforming activity. - ANSWER D. DNase destroyed the
transforming activity.
(Treatment of the transforming principle with DNase destroyed the DNA and thus its ability to transform
bacteria.)
, (T/F) Guanine and adenine are purines found in DNA. - ANSWER 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. Nucleic acids are formed through phosphodiester bonds that link nucleosides together.
B. The pentose sugar in DNA is ribose.
C. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in
opposite directions.
D. Hydrogen bonds formed between the sugar-phosphate backbones of the two DNA chains help to
stabilize DNA structure. - ANSWER C. 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′? - ANSWER 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. a cesium chloride gradient
B. a means of distinguishing among the distribution patterns of newly synthesized and parent molecule
DNA possible
C. that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules
D. the fact that DNA is the genetic material - ANSWER D. the fact that DNA is the genetic
material
(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 _______. - ANSWER
distinguish between semi-conservative and dispersive replication.
(After one round of replication, the results of these two possibilities are indistinguishable. A second round
was required to distinguish between these two possibilities.)
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 heavy band
D. One light band - ANSWER B. One light band and one intermediate band.
(Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.)
Answers With Verified Solutions Graded A+ Latest
Update 2025.
All EXCEPT which of the following are characteristics of the genetic material?
A. It must be replicated accurately.
B. It contains all the information needed for growth, development, and reproduction of the organism.
C. It must be capable of change.
D. It is composed of protein. - ANSWER 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. Each scientist had his own method for labeling phage, so each conducted the same experiment using a
different isotope.
B. Establishing the identity of the genetic material required observation of two phage generations.
C. The bacteriophage used in the experiments was a T2 phage.
D. It was necessary that each of the two phage components, DNA and protein, be identifiable upon
recovery at the end of the experiment. - ANSWER D. 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. Phage T2 is capable of replicating within a bacterial host.
B. DNA is the identity of the hereditary material in phage T2.
C. Some viruses can infect bacteria.
,D. When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates
radioactive label. - ANSWER B. 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. The phage would fail to infect bacteria.
B. Neither preparation of infected bacteria would exhibit radioactivity.
C. Both preparations of infected bacteria would exhibit radioactivity.
D. Both preparations of infected bacteria would contain both P32 and S35. - ANSWER 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. That DNA is the genetic material.
B. The heat-killed, virulent Streptococcus pneumoniae was lethal to the mouse.
C. 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 survived injection of live virulent (smooth) Streptococcus pneumoniae. - ANSWER
C. 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. Protease destroyed the transforming activity.
B. The transforming principle was too complex and difficult to be purified.
C. RNase destroyed the transforming activity.
D. DNase destroyed the transforming activity. - ANSWER D. DNase destroyed the
transforming activity.
(Treatment of the transforming principle with DNase destroyed the DNA and thus its ability to transform
bacteria.)
, (T/F) Guanine and adenine are purines found in DNA. - ANSWER 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. Nucleic acids are formed through phosphodiester bonds that link nucleosides together.
B. The pentose sugar in DNA is ribose.
C. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in
opposite directions.
D. Hydrogen bonds formed between the sugar-phosphate backbones of the two DNA chains help to
stabilize DNA structure. - ANSWER C. 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′? - ANSWER 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. a cesium chloride gradient
B. a means of distinguishing among the distribution patterns of newly synthesized and parent molecule
DNA possible
C. that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules
D. the fact that DNA is the genetic material - ANSWER D. the fact that DNA is the genetic
material
(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 _______. - ANSWER
distinguish between semi-conservative and dispersive replication.
(After one round of replication, the results of these two possibilities are indistinguishable. A second round
was required to distinguish between these two possibilities.)
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 heavy band
D. One light band - ANSWER B. One light band and one intermediate band.
(Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.)
